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d & f – block – JEE Main Previous Year Questions with Solutions
JEE Main Previous Year Papers Questions of Chemistry with Solutions are available at eSaral. Practicing JEE Main chapter wise questions of Chemistry will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.SimulatorPrevious Years AIEEE/JEE Main Question
Q. In context with the transition elements, which of the following statements is incorrect?(1) In the highest oxidation states of the first five transition elements (Sc to Mn), all the 4sand 3d electrons are used for bonding.(2) Once the d5 configuration is exceeded, the tendency to involve all the 3d electrons inbonding decreases.(3) In addition to the normal oxidation states, the zero oxidation state is also shown by theseelements in complexes.(4) In the highest oxidation states, the transition metal show basic character and form cationiccomplexes. [AIEEE-2009]

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Sol. (4)

Q. Iron exhibits +2 and +3 oxidation states. Which of the following statements about iron is incorrect ?(1) Ferrous compounds are more easily hydrolysed than the corresponding ferric compounds.(2) Ferrous oxide is more basic in nature than the ferric oxide.(3) Ferrous compounds are relatively more ionic than the corresponding ferric compounds.(4) Ferrous compounds are less volatile than the corresponding ferric compounds. [AIEEE-2012]

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Sol. (1)

Q. Consider the following reaction : $\mathrm{xMnO}_{4}^{-}+\mathrm{yC}_{2} \mathrm{O}_{4}^{2-}+\mathrm{zH}^{+} \rightarrow \mathrm{xMn}^{2+}+2 \mathrm{yCO}_{2}+\frac{\mathrm{Z}}{2} \mathrm{H}_{2} \mathrm{O}$The values of x, y and z in the reaction are respectively :-(1) 5,2 and 16(2) 2,5 and 8(3) 2, 5 and 16(4) 5,2 and 8 [JEE MAIN-2013]

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Sol. (3)

Q. Which of the following arrangements does not represent the correct order of the property stated against it ?(1) $\mathrm{V}^{2+}<\mathrm{Cr}^{2+}<\mathrm{Mn}^{2}<\mathrm{Fe}^{2+}$: paramagnetic behaviour(2) $\mathrm{Ni}^{2+}<\mathrm{Co}^{2+}<\mathrm{Fe}^{2+}<\mathrm{Mn}^{2+}$: ionic size(3)$\mathrm{Co}^{3+}<\mathrm{Fe}^{3+}<\mathrm{Cr}^{3+}<\mathrm{Sc}^{3+}$: stability in aqueous solution(4) $\mathrm{Sc}<\mathrm{Ti}<\mathrm{Cr}<$Mn : number of oxidation states [JEE MAIN-2013]

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Sol. (1)

Q. Potassium dichromate when heated with concentrated sulphuric acid and a soluble chloride, gives brown – red vapours of:(1) $\mathrm{CrO}_{3}$(2) $\mathrm{Cr}_{2} \mathrm{O}_{3}$(3) $\mathrm{CrCl}_{3}$(4) $\mathrm{CrO}_{2} \mathrm{Cl}_{2}$ [JEE MAIN-2013, Online]

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Sol. (4)Explanation. $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}+6 \mathrm{H}_{2} \mathrm{SO}_{4}+4 \mathrm{NaCl} \longrightarrow 2 \mathrm{KHSO}_{4}+4 \mathrm{NaHSO}_{4}+2 \mathrm{CrO}_{2} \mathrm{Cl}_{2}+3 \mathrm{H}_{2} \mathrm{O}$

Q. The element with which of the following outer electron configuration may exhibit the largest number of oxidation states in its compounds :(1) $3 \mathrm{d}^{7} 4 \mathrm{s}^{2}$(2) $3 \mathrm{d}^{8} 4 \mathrm{s}^{2}$( 3) (3) $3 \mathrm{d}^{5} 4 \mathrm{s}^{2}$(4) (4) $3 \mathrm{d}^{6} 4 \mathrm{S}^{2}$ [JEE MAIN-2013, Online]

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Sol. (3)Explanation$\mathrm{Mn} \longrightarrow \mathrm{Mn}^{+7}$

Q. When a small amount of $\mathrm{KMnO}_{4}$ is added to concentrated $\mathrm{H}_{2} \mathrm{SO}_{4}$a green oily compound is obtained which is highly explosive in nature. Compound may be:(1) $\mathrm{Mn}_{2} \mathrm{O}_{3}$(2) MnSO $_{4}$(3) $\mathrm{Mn}_{2} \mathrm{O}_{7}$( 4) $\mathrm{MnO}_{2}$ [JEE MAIN-2013, Online]

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Sol. (3)Explanation

Q. Which series of reactions correctly represents chemical relations related to iron and its compound? [JEE MAIN-2014]

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Sol. (2)

Q. The equation which is balanced and represents the correct product (s) is: [JEE MAIN-2014]

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Sol. (4)

Q. Which of the following is not formed when $\mathrm{H}_{2} \mathrm{S}$ reacts with acidic $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$solution ?( 1) $\mathrm{K}_{2} \mathrm{SO}_{4}$(2) $\mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}$(3) $\mathrm{S}$(4) $\mathrm{CrSO}_{4}$ [JEE MAIN-2014, Online]

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Sol. (4)

Q. Copper becomes green when exposed to moist air for a long period. This is due to :-(1) the formation of a layer of cupric oxide on the surface of copper.(2) the formation of basic copper sulphate layer on the surface of the metal(3) the formation of a layer of cupric hydroxide on the surface of copper.(4) the formation of a layer of basic carbonate of copper on the surface of copper. [JEE MAIN-2014, Online]

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Sol. (4)

Q. Which one of the following exhibits the largest number of oxidation states ?(1) Mn(25)(2) V(23)(3) Cr (24)(4) Ti (22) [JEE MAIN-2014, Online]

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Sol. (1)

Q. How many electrons are involved in the following redox reaction?$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+\mathrm{Fe}^{2+}+\mathrm{C}_{2} \mathrm{O}_{4}^{2-} \rightarrow \mathrm{Cr}^{3+}+\mathrm{Fe}^{3+}+\mathrm{CO}_{2}$ (Unbalanced)(1) 3(2) 4(3) 5(4) 6 [JEE MAINS-2014,Online]

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Sol. (4)

Q. Amongst the following, identify the species with an atom in +6 oxidation state:(1) $\left[\mathrm{MnO}_{4}\right]^{-}$(2) $\left[\mathrm{Cr}(\mathrm{CN})_{6}\right]^{3-}$(3) $\mathrm{Cr}_{2} \mathrm{O}_{3}$(4) $\mathrm{CrO}_{2} \mathrm{Cl}_{2}$ [JEE MAIN-2014, Online]

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Sol. (4)

Q. Match the catalysts to the correct processes :-(1) A-ii, B-iii, C-iv, D-i(2) A-iii, B-i, C-ii, D-iv(3) A-iii, B-ii, C-iv, D-i(4) A-ii, B-i, C-iv, D-iii [JEE MAIN-2015]

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Sol. (4)

Q. Which of the following statements is false :-(1) has a Cr – O – Cr bond(2) is tetrahedral in shape(3) $\mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ is a primary standard in volumetry(4) $\mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ is less soluble than $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ [JEE MAIN-2015, Online]

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Sol. (3)

Polymer – JEE Main Previous Year Questions with Solutions
JEE Main Previous Year Papers Questions of Chemistry with Solutions are available at eSaral. Practicing JEE Main chapter wise questions of Chemistry will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.  Simulator Previous Years AIEEE/JEE Main Questions
Q. Buna–N synthetic rubber is a copolymer of :- [AIEEE-2009]

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Sol. (1)Buna-N is actually abbrebiated form, where Bu represents 1, 3-butadiene, Na representssodium and N represents nitrile (acrylonitrile).Thus, Buna-N is a copolymer of 1, 3-butadiene and acrylonitrile usually copolymer of1, 3-butadiene, and acrylonitrile usually ploymerise in the presence of sodium.

Q. The polymer containing strong intermolecular forces e.g. hydrogen bonding, is :-(1) natural rubber(2) teflon(3) nylon 6, 6(4) polystyrene [AIEEE-2010]

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Sol. (3)

Q. Thermosetting polymer, Bakelite is formed by the reaction of phenol with :-(1) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHO}$(2) $\mathrm{CH}_{3} \mathrm{CHO}$(3) HCHO(4) HCOOH [AIEEE-2011]

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Sol. (3)Bakelite is a thermosetting polymer formed by the condensation reaction of phenol withHCHO in the presense of conc. $\mathrm{H}_{2} \mathrm{SO}_{4}$It is thus, a cross-linked polymer in which condesation takes place at O- and p- positionsSo HCHO is the corerect option.

Q. The species which can best serve as an initiator for the cationic polymerization is :-(1) BuLi(3) HNO $_{3}$(4) $\mathrm{AlCl}_{3}$ [AIEEE-2012]

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Sol. (4)Electron – deficient species (Lawis acid) like $\mathrm{AlCl}_{3}, \mathrm{BF}_{3}$ etc.are is used as initiator for cationic polymerisation.

Q. Which one is classified as a condensation polymer?(1) Teflon(2) Acrylonitrile(3) Dacron(4) Neoprene [JEE-Main 2014]

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Sol. (3)

Q. Which one of the following structures represents the neoprene polymer :- [JEE-Main(Online) – 2015]

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Sol. (3)

Q. Which polymer is used in the manufacture of paints and lacquers ?(1) Polypropene(2) Poly vinyl chloride(3) Bakelite(4) Glyptal [JEE- Main – 2015]

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Sol. (4)

Q. Which of the following statements about low density polythene is FALSE ?(1) It is used in the manufacture of buckets, dust-bins etc.(2) Its synthesis requires high pressure(3) It is a poor conductor of electricity(4) Its synthesis requires dioxygen or a peroxide initiator as a catalyst. [JEE Main 2016]

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Sol. (1)Low density polythene : It is obtained by the polymerisation of ethene under high presure of 1000-2000 atm. at a temp. of 350 K to 570 K in the pressure of traces of dioxygen or a peroxide

Q. Assertion : Rayon is a semisynthetic polymer whose properties are better thannatural cotton.Reason : Mechanical and aesthetic properties of cellulose can be improved by acetylation.(1) Both assertion and reason are incorrect(2) Assertion is incorrect statement, but the reason is correct(3) Both assertion and reason are correct, but the reason is not the correct explanation for theassertion(4) Both assertion and reason are correct, and the reason is the correct explanation for the assertion [JEE Main (Online) 2016]

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Sol. (4)

Q. Which of the following polymers is synthesized using a free radical polymerizationtechnique ? (1) Terylene(2) Teflon(3) Nylon 6,6(4) Melamine polymer [JEE Main (Online) 2016]

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Sol. (2)[esquestion] The formation of which of the following polymers involves hydrolysis reaction ?(1) Nylon 6(2) Bakelite(3) Nylon 6, 6(4) Terylene#tag# [JEE – Main 2017]

Chemistry in Everyday Life – JEE Main Previous Year Questions with Solutions
JEE Main Previous Year Papers Questions of Chemistry with Solutions are available at eSaral. Practicing JEE Mains chapter wise questions of Chemistry will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.SimulatorPrevious Years AIEEE/JEE Mains Questions
Q. Which artificial sweetener contains chlorine ?(1) Aspartame(2) Saccharin(3) Sucralose(4) Alitame [JEE-Main(Online) – 2015]

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Sol. (3)

Q. Thiol group is present in :-(1)Methionine(2) Cytosine(3) Cystine(4) Cysteine [JEE-Main-2016]

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Sol. (4)Among 20 naturally occuring amino acids “Cysteine” has ‘– SH’ or thiol functional group.

Q. The distillation technique most suited for separating glycerol from spent-lye in the soap industry is:(1) Distillation under reduced pressure(2) Simple distillation(3) Fractional distillation(4) Steam distillation (Jee-Main-2016)

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Sol. (1)(1) Distillation under reduced pressure.Glycerol $\left(\mathrm{B} . \mathrm{P} .290^{\circ} \mathrm{C}\right)$ is separated from spent lye in the soap industry by distillation under reduced pressure, as for simple distillation very high temperature is required which might decompose the component.

Q. The predominant form of histamine present in human blood is $\left(\mathrm{pK}_{\mathrm{a}}\right.$, Histidine = 6.0)(1) (2) (3) (4)

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Sol. (3)

Carboxylic Acid – JEE Main Previous Year Questions with Solutions
JEE Main Previous Year Papers Questions of Chemistry with Solutions are available at eSaral. Practicing JEE Mains chapter wise questions of Chemistry will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.SimulatorPrevious Years AIEEE/JEE Mains Questions
Q. In the chemical reactions,(1) Nitrobenzene and chlorobenzene(2) Nitrobenzene and fluorobenzene(3) Phenol and benzene(4) Benzene diazonium chloride and fluorobenzene [AIEEE-2010]

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Sol. (4)

Q. In the chemical reactions(1) Fluorobenzene and phenol(2)Benzene diazonium chloride and benzonitrile(3) Nitrobenzene and chlorobenzene(4) Phenol and bromobenzene [AIEEE-2011]

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Sol. (2)

Q. Compound (A), $\mathrm{C}_{8} \mathrm{H}_{9} \mathrm{Br}$, gives a white precipitate when warmed with alcoholic $\mathrm{AgNO}_{3}$. Oxidation of (A) gives an acid (B), $\mathrm{C}_{8} \mathrm{H}_{6} \mathrm{O}_{4}$. (B) easily forms anhydride on heating. Identify the compound (A): [AIEEE-2013]

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Sol. (4) since B on heating gives anhydride so B is 1,2 dicarboxylic acid. It contains two Benzylic carbon at 1,2 position so on oxidation it produces 1, 2-dicarboxylic acid

Q. An organic compound A upon reacting with $\mathrm{NH}_{3}$ gives B. On heating, B gives C. C in presence of KOH reacts with $\mathbf{B} \mathbf{r}_{2}$ to give $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{2}$. A is :-(1) $\mathrm{CH}_{3} \mathrm{COOH}$(2) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COOH}$(4) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}$ [AIEEE-2013]

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Sol. (4)

Q. On heating an aliphatic primary amine with chloroform & ethenolic potassium hydroside the organic compound formed is(1) An alkyl cyanide(2) An alkyl isocyanide(3) an alkanol(4) an alkanediol [AIEEE-2014]

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Sol. (2)It is a carbyl amine reaction used for identification of primary amine also known as isocynide test because of offensive smell of isocynide.

Q. In the reaction(1) Ethylene(2) Acetyl chloride(3) Acetaldehyde(4) Acetylene. [AIEEE-2014]

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Sol. (1)

Q. In the presence of a small amount of phosphorous, aliphatic carboxylic acids react with chlorine or bromine to yield a compound in which  – hydrogen has been replaced by halogen. This reaction is known as :(1) Etard reaction(2) Hell – Volhard – Zelinsky reaction(3) Wolff – Kischner reaction(4) Rosenmund reaction [JEE(Main)-2015]

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Sol. (2)

Q. In the Hofmann bromamide degradation reaction, the number of moles of NaOH and $\mathrm{Br}_{2}$ used per mole of amine produced are : [JEE(Main)-2016]

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Sol. (1)

Q. The major product obtained in the following reaction is :- [JEE – Main 2017]

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Sol. (1)DIBAL – H is electrophilic reducing agent reduces cynide, esters, lactone, amide, carboxylic acid into corresponding Aldehyde (partial reduction)

Q. Sodium salt of an organic acid ‘X’ produces effervescence with conc. $\mathrm{H}_{2} \mathrm{SO}_{4}$. ‘X’ reacts with the acidified aqueous CaCl2 solution to give a white precipitate which decolourises acidic solution of $\mathrm{KMnO}_{4}$. ‘X’ is :-(1) $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COONa}$(2) HCOONa(3) CH3COONa(4) $\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}$ [JEE – Main – 2017]

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Sol. (4)

Q. Which of the following compounds will be suitable for Kjeldahl’s method for nitrogen estimation ? [JEE(Main)-2018]

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Sol. (1)Kjeldahl’s is suitable for Aniline. This method is used for quantitative analysis of N compound in organic substance $\left(\mathrm{NH}_{3} / \mathrm{NH}_{4}^{+}\right) .$

Carbonyl Compounds – JEE Main Previous Year Questions with Solutions
JEE Main Previous Year Papers Questions of Chemistry with Solutions are available at eSaral. Practicing JEE Mains chapter wise questions of Chemistry will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.Simulator Previous Years AIEEE/JEE Mains Questions
Q. Which of the following on heating with aqueous KOH, produces acetaldehyde ?(1) $\mathrm{CH}_{2} \mathrm{ClCH}_{2} \mathrm{Cl}$(2) $\mathrm{CH}_{3} \mathrm{CHCl}_{2}$(3) $\mathrm{CH}_{3} \mathrm{COCl}$(4) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}$ [AIEEE-2009]

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Sol. (2)

Q. In Cannizzaro reaction given below :-$2 \mathrm{Ph} \mathrm{CHO}$ $\mathrm{PhCH}_{2} \mathrm{OH}+\mathrm{PhCO}_{2}^{\Theta}$ the slowest step is :-(1) The abstraction of proton from the carboxylic group(2) The deprotonation of $\mathrm{CH}_{2} \mathrm{OH}$(3) The attack of $: \stackrel{\ominus}{\mathrm{O}} \mathrm{H}$ at the carboxyl group(4) The transfer of hydride to the carbonyl group [AIEEE-2009]

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Sol. (4)

Q. One mole of a symmetrical alkene on ozonolysis gives two moles of an aldehyde having a molecular mass of 44 u. The alkene is :-(1) Ethene (2) Propene (3) 1-Butene (4) 2-Butene [AIEEE-2010]

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Sol. (4)

Q. Ozonolysis of an organic compound gives formaldehyde as one of the products. This confirms the presence of :-(1) An isopropyl group(2) An acetylenic triple bond(3) Two ethylenic double bonds(4) A vinyl group [AIEEE-2011]

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Sol. (4)

Q. Ozonolysis of an organic compound ‘A’ produces acetone and propionaldehyde in equimolar mixture. Identify ‘A’ from the following compounds : –(1) 2-Methyl – 1- pentene(2) 1-Pentene(3) 2-Pentene(4) 2-Methyl-2-pentene [AIEEE-2011]

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Sol. (4)

Q. Trichloroacetaldehyde was subjected to Cannizzaro’s reaction by using NaOH. The mixture of the products contains sodium trichloroacetate and another compound. The other(1) 2,2,2–Trichloropropanol(2) Chloroform(3) 2,2,2–Trichloroethanol(4) Trichloromethanol [AIEEE-2011]

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Sol. (3)

Q. Silver Mirror test is given by which one of the following compounds?(1) Formaldehyde (2) Benzophenone (3) Acetaldehyde (4) Acetone [AIEEE-2011]

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Sol. (1,3)

Q. In the given transformation, which of the following is the most appropriate reagent ?(1) $\mathrm{NaBH}_{4}$(2) $\mathrm{NH}_{2} \mathrm{NH}_{2}, \stackrel{\ominus}{\mathrm{O}} \mathrm{H}$(3) Zn – Hg / HCl$(4) \mathrm{Na}, \mathrm{Liq.NH}_{3}$ [AIEEE-2012]

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Sol. (2)

Q. Iodoform can be prepared from all except :-(1) Isobutyl alcohol(2) Ethyl methyl ketone(3) Isopropyl alcohol(4) 3-Methyl–2–butanone [AIEEE-2012]

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Sol. (1)

Q. A compound with molecular mass 180 is acylated with $\mathrm{CH}_{3} \mathrm{COCl}$ to get a compound with molecular mass 390. The number of amino groups present per molecule of the former compound is :-(1) 2 (2) 5 (3) 4 (4) 6 [JEE(Main)-2013]

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Sol. (2)180 molecular mass after acylation gives compound of molecular mass 390.Change in molecular mass = 390 – 180 = 210so number of $\mathrm{NH}_{2}$= $\frac{210}{42}$ = 5 group

Q. The major organic compound formed by the reaction of 1, 1, 1–trichloroethane with silver powder is :-(1)2-Butyne(2) 2-Butene(3) Acetylen(4) Ethene [JEE(Main)-2014]

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Sol. (1)

Q. The most suitable reagent for the conversion of $\mathrm{R}-\mathrm{CH}_{2}-\mathrm{OH} \rightarrow \mathrm{R}-\mathrm{CHO}$ is :-(1) $\mathrm{CrO}_{3}$(2) PCC (Pyridinium chlorochromate)(3) $\mathrm{KMNO}_{4}$(4) $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ [JEE(Main)-2014]

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Sol. (2)PCC is milder oxidising agent, oxidises only into aldehyde.

Q. A compound A with molecular formula $\mathrm{C}_{10} \mathrm{H}_{13} \mathrm{Cl}$ gives a white precipitate on adding silver nitrate solution. A on reacting with alcoholic KOH gives compound B as the main product. B on ozonolysis gives C and D. C gives Cannizaro reaction but not aldol condensation. D gives aldol condensation but not Cannizaro reaction. A is :#tag [JEE(Main)-2015]

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Sol. (2)

Q. In the reaction sequence ; the product B is:-(1) $\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}-\mathrm{CHO}$(2) $\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{3}$(3) $\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{OH}$(4) [JEE(Main)-2015]

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Sol. (1)

Q. Which compound would give 5-keto-2-methyl hexanal upon ozonlysis? [JEE(Main) 2015]

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Sol. (4)

Biomolecule – JEE Main Previous Year Questions with Solutions
JEE Main Previous Year Papers Questions of Chemistry with Solutions are available at eSaral. Practicing JEE Mains chapter wise questions of Chemistry will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.Simulator Previous Years AIEEE/JEE Main Questions
Q. The two functional groups present in a typical carbohydrate are :-(1) >C = O and –OH (2) –OH and –CHO (3) –OH and –COOH (4) –CHO and –COOH AIEEE-2009

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Sol. (2)

Q. Biurest test is not given by :-(1) proteins (2) carbohydrates (3) polypeptides (4) urea AIEEE-2010

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Sol. (2)

Q. The presence or absence of hydroxy group on which carbon atom of sugar differentiates RNA and DNA?(1) 3rd (2) 4th (3) 1st (4) 2nd AIEEE-2011

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Sol. (4)

Q. The change in the optical rotation of freshly prepared solution of glucose is known as :-(1) tautomerism (2) racemisation (3) specific rotation (4) mutarotation AIEEE-2011

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Sol. (4)

Q. Which one of the following statements is correct ?(1) All amino acids except glutamic acid are optically active(2) All amino acids except lysine are optically active(3) All amino acids are optically active(4) All amino acids except glycine are optically active AIEEE-2012

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Sol. (4)

Q. Synthesis of each molecule of glucose in photosynthesis involves :-(1) 18 molecules of ATP(2) 10 molecules of ATP(3) 8 molecules of ATP(4) 6 molecules of ATP JEE Main -2013

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Sol. (1)Six rounds of the Calvin cycle are required, because one carbon atom is reduced in each round. Twelve molecules of ATP are expended. An additional six molecular of ATP are spent in regenerating ribulose-1, 5-biphosphate$6 \mathrm{CO}_{2}+18 \mathrm{ATP}+12 \mathrm{NADPH}+12 \mathrm{H}_{2} \mathrm{O} \rightarrow$$\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}+18 \mathrm{ADP}+18 \mathrm{P}_{\mathrm{i}}+12 \mathrm{NADP}+6 \mathrm{H} \oplusReference : NCERT 11th class chapter-13 (Photosynthesis Pg. 217, 218 in higher plants.Calvin cycle where sugar is synthesised is as follows :The Calvin cycle proceeds in three stages (1) carboxylation, during which \mathrm{CO}_{2} combines with ribulose-1.5-bisphosphate (2) reduction, during which carbohydrate is formed at the expense of the photochemically made ATP and NADPH and (3) regeneration during which the \mathrm{CO}_{2} acceptor ribulose-1,5-bisphosphate is formed again so that the cylcle continues.It might help you to understand all of this if we look at what goes in and what comes out of the Calvin cycle. Q. Which one of the following bases is not present in DNA ?(1) Cytosine (2) Thymine (3) Quinoline (4) Adenine JEE-Main 2014 Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (3)All cytosine, thymine and adenine are present in DNA. Only quinoline is not present in DNA. Q. Which of the vitamins given below is water soluble?(1) Vitamin E(2) Vitamin K(3) Vitamin C(4) Vitamin D JEE-Main 2015 Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (3)Vitamine C is soluble in water due to H-Bonding & all other vitamines given in question are fat soluble. Q. Complete hydrolysis of starch gives :(1) glucose and fructose in equimolar amounts(2) glucose only(3) galactose and fructose in equimolar amounts(4) glucose and galactose in equimolar amounts JEE-Main(Online) – 2015 Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (2)Starch is polymer of Glucose Q. Accumulation of which of the following molecules in the muscles occurs as a result of vigorous exercise :-(1) Pyruvic acid(2) L-lactic acid(3) Glycogen(4) Glucose JEE-Main(Online) – 2015 Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (2) Q. Consider the following sequence for aspartic acid:The pI (isoelectric point) of aspartic acid is :(1) 1.88 (2) 2.77 (3) 3.65 (4) 5.74 JEE-Main(Online) – 2016 Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (4) Q. Observation of “Rhumann’s purple” is a confirmatory test for the presence of :(1) Starch(2) Reducing sugar(3) Cupric ion(4) Protein JEE-Main(Online) – 2017 Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (1) Q. Which of the following compounds will behave as a reducing sugar in an aqueous KOH solution ? JEE-Main 2017 Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (1)(1) Ester in presence of Aqueous KOH solution give SNAE reaction so following reaction takes place Alcohol & Ether – JEE Main Previous Year Questions with Solutions JEE Main Previous Year Papers Questions of Chemistry with Solutions are available at eSaral. Practicing JEE Mains chapter wise questions of Chemistry will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.SimulatorPrevious Years AIEEE/JEE Mains Questions Q. The number of stereoisomers possible for a compound of the molecular formula(1) 4 (2) 6 (3) 3 (4) 2 AIEEE-2009 Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (1) Q. A liquid was mixed with ethanol and a drop of concentrated \mathrm{H}_{2} \mathrm{SO}_{4} was added. A compound with a fruity smell was formed. The liquid was :-(1) \mathrm{CH}_{3} \mathrm{COCH}_{3}(2) \mathrm{CH}_{3} \mathrm{COOH}(3) \mathrm{CH}_{3} \mathrm{OH}(4) HCHO AIEEE-2009 Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (2) Q. From amongst the following alcohols the one that would react fastest with conc. HCl and anhydrous \mathrm{ZnCl}_{2}, is :-(1) 1–Butanol(2) 2–Butanol(3) 2–Methylpropan–2–ol(4) 2–Methylpropanol AIEEE-2010 Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (3) Q. Consider the following reaction :\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow ProduceAmong the following, which one cannot be formed as a product under any conditions ?(1) Ethyl-hydrogen sulphate(2) Ethylene(3) Acetylene(3) Diethyl ethe AIEEE-2011 Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (3) Q. Iodoform can be prepared from all except :-(1) Isobutyl alcohol(2) Ethyl methyl ketone(3) Isopropyl alcohol(4) 3-Methyl–2–butanone AIEEE-2012 Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (1) Q. An unknown alcohol is treated with the “Lucas reagent’ to determine whether the alcohol is primary, secondary or tertiary. Which alcohol reacts fastest and by what mechanism :-(1) secondary alcoholby SN ^{1}(2) tertiary alcohol by SN ^{1}(3) secondary alcoholby SN ^{2}(4) tertiary alcohol by SN ^{2} AIEEE-2013 Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (2) Q. The most suitable reagent for the conversion of \mathrm{R}-\mathrm{CH}_{2}-\mathrm{OH} \rightarrow \mathrm{R}-\mathrm{CHO} is :(1) \mathrm{CiO}_{3}(2) PCC (Pyridinium chlorochromate)(3) KMNO _{4}(4) \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} Jee-Main-2014 Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (2) Q. Allyl phenyl ether can be prepared by heating:(1) \mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2}-\mathrm{Br}+\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{ONa}(2) \mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{CH}=\mathrm{CH}-\mathrm{Br}+\mathrm{CH}_{3}-\mathrm{ONa}(3) \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Br}+\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2}-\mathrm{ONa}(4) \mathrm{CH}_{2}=\mathrm{CH}-\mathrm{Br}+\mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{CH}_{2}-\mathrm{ONa} Jee-Main-2014 Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (1) Q. In the Victor-Meyer’s test, the colour given by 1^{\circ}, 2^{\circ} and 3^{\circ} alcohols are respectively :-(1) Red, blue, colourless(2) Colourless, red, blue(3) Red, blue, violet(4) Red, colourless, blue Jee-Main-2014 Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (1) Q. Williamson synthesis of ether is an example of(1) Nucleophilic addition(2) Electrophilic substitution(3) Nucleophilic substitution(4) Electrophilic addition Jee-Main-2014 Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (3)Nucleophilic substitution Q. Phenol on treatment with \mathrm{CO}_{2} in the presence of NaOH followed by acidification produces compound X as the major product. X on treatment with \left(\mathrm{CH}_{3} \mathrm{CO}\right)_{2} \mathrm{O} in the presence of catalytic amount of \mathrm{H}_{2} \mathrm{SO}_{4} produces : Jee-Main-2018 Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (4) Q. Phenol reacts with methyl chloroformate in the presence of NaOH to form product A. A reacts with \mathrm{Br}_{2} to form product B. A and B are respectively : Jee-Main-2018 Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (2) Salt Analysis – JEE Main Previous Year Questions with Solutions JEE Main Previous Year Papers Questions of Chemistry with Solutions are available at eSaral. Practicing JEE Main chapter wise questions of Chemistry will help the JEE aspirants in realising the question pattern as well as help in analysing weak & strong areas. Simulator Previous Years AIEEE/JEE Mains Questions Q. The products obtained on heating LiNO3 will be :-(1) \mathrm{LiNO}_{2}+\mathrm{O}_{2}(2) \mathrm{Li}_{2} \mathrm{O}+\mathrm{NO}_{2}+\mathrm{O}_{2}(3) \mathrm{Li}_{3} \mathrm{N}+\mathrm{O}_{2}(4) \mathrm{Li}_{2} \mathrm{O}+\mathrm{NO}+\mathrm{O}_{2} AIEEE-2011 Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (2)\mathrm{LiNO}_{3} \longrightarrow \mathrm{Li}_{2} \mathrm{O}+\mathrm{NO}_{2}+\mathrm{O} Q. What is the best description of the change that occurs when \mathrm{Na}_{2} \mathrm{O}(\mathrm{s}) is dissolved in water ?(1) Oxidation number of sodium decreases(2) Oxide ion accepts sharing in a pair of electrons(3) Oxide ion donates a pair of electrons(4) Oxidation number of oxygen increases AIEEE-2011 Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (3)\mathrm{Na}_{2} \mathrm{O}+\mathrm{HOH} \longrightarrow \mathrm{CaO}+\mathrm{CO}_{2} Q. Which of the following on thermal-decomposition yields a basic as well as an acidi oxide ?(1) \mathrm{NH}_{4} \mathrm{NO}_{3}(2) NaNO _{3}(3) \mathrm{KClO}_{3}(4) \mathrm{CaCO}_{3} AIEEE-2012 Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (4)\mathrm{CaCO}_{3} \longrightarrow \mathrm{CaO}+\mathrm{CO}_{2} Q. Fire extinguishers contain H2SO4 and which one of the following :-(1) \mathrm{CaCO}_{3}(2) \mathrm{NaHCO}_{3} and \mathrm{Na}_{2} \mathrm{CO}_{3}(3) \mathrm{Na}_{2} \mathrm{CO}_{3}(4) \mathrm{NaHCO}_{3} JEE(Main)-2012 online-P-1 Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (4)Fire extinguishers contain H2SO4 and NaHCO3 Q. Copper wire test for halogens is known as :-(1) Duma’s Test(2) Beilstein’s Test(3) Lasssigne’s Test(4) Liebig’s Test JEE(Main)-2012 online-P-2 Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (2)Beilstein’s Test Q. The standard potentials of \mathrm{Ag}^{+} / \mathrm{Ag}, \mathrm{Hg}_{2}^{2+} / 2 \mathrm{Hg}, \mathrm{Cu}^{2+} / \mathrm{Cu} and \mathrm{Mg}^{2+} / Mg electrodes are 0.80, 0.79, 0.34 and – 2.37 V, respectively. An aqueous solution which contains one mole per litre of the salts of each of the four metals is electrolyzed. With increasing voltage, the correct sequence of deposition of the metals at the cathode is :-(1) Cu, Hg, Ag only(2) Mg, Cu, Hg, Ag(3) Ag, Hg, Cu only(4) Ag, Hg, Cu, Mg [JEE(Main)-2012 online-P-3] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (3) Q. Beilstein test is used for estimation of which one of following elements ?(1) S(2) Cl(3) C and H(4) N JEE(Main)-2012 online-P-3 Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (2)5 \mathrm{Mn} \mathrm{O}_{4}^{-}+2 \mathrm{C}_{2} \mathrm{O}_{4}^{2-}+16 \mathrm{H}^{+} \longrightarrow 5 \mathrm{Mn}^{+2}+2 \mathrm{CO}_{2}+\frac{16}{2} \mathrm{H}_{2} \mathrm{O} Q. In the following balanced reaction, \mathrm{XMnO}_{+}^{-}+\mathrm{YC}_{2} \mathrm{O}_{4}^{2-}+\mathrm{ZH}^{+}=\mathrm{XMn}^{2+}+2 \mathrm{YCO}_{2}+ \frac{\mathrm{Z}}{2} \mathrm{H}_{2} \mathrm{O} the values of X, Y and Z, respectively are :- JEE(Main)-2012 online-P-4 Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (2) Q. A metal M on heating in nitrogen gas gives Y. Y on treatment with \mathrm{H}_{2} \mathrm{O} gives a colourless gas which when passed through \mathrm{CuSO}_{4} solution gives a blue colour, Y is :-(1) \mathrm{NH}_{3}(2) MgO(3) \mathrm{Mg}_{3} \mathrm{N}_{2}(4) \mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2} JEE(Main)-2012 online-P-4 Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (3) Q. Electrode potentials (E0) are given below :\mathrm{Cu}^{+} / \mathrm{Cu}=+0.52 \mathrm{V}, \mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}=+0.77 \mathrm{V}$$1 / 2 \mathrm{I}_{2}(\mathrm{s}) / \mathrm{I}^{-}=+0.54 \mathrm{V}, \mathrm{Ag}^{+} / \mathrm{Ag}=+0.88 \mathrm{V}$Based on the above potentials, strnogest oxidizing agent will be :(1) $\mathrm{Cu}^{+}$(3) $\mathrm{Ag}^{+}$(4) $\mathrm{I}_{2}$ JEE(Main)-2013 online-P-1

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Sol. (3)

Q. Potassium dichromate when heated with concentrated sulphuric acid and a soluble chloride, gives brown – red vapours of:( 1) $\mathrm{CrO}_{3}$(2) $\mathrm{Cr}_{2} \mathrm{O}_{3}$(3) $\mathrm{CrCl}_{3}$(4) $\mathrm{CrO}_{2} \mathrm{Cl}_{2}$ JEE(Main)-2013 online-P-1

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Sol. (4)

Q. Given :$\overline{\mathrm{X}} \mathrm{Na}_{2} \mathrm{HAsO}_{3}+\mathrm{Y} \mathrm{NaBrO}_{3}+\mathrm{ZHCl} \rightarrow \mathrm{NaBr}+\mathrm{H}_{3} \mathrm{AsO}_{4}+\mathrm{NaCl}$The values of X, Y and Z in the above redox reaction are respectively :(1) 2, 1, 3 (2) 3, 1, 6 (3) 2, 1, 2 (4) 3, 1, 4 JEE(Main)-2013 online-P-1

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Sol. (2)$3 \mathrm{Na}_{2} \mathrm{HAsO}_{3}+\mathrm{NaBrO}_{3}+6 \mathrm{HCl} \longrightarrow \mathrm{NaBr}+\mathrm{H}_{3} \mathrm{AsO}_{4}+\mathrm{NaCl}$

Q. Sodium Carbonate cannot be used in place of $\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}$ for the identification of $\mathrm{Ca}^{2+}, \mathrm{Ba}^{2+}$ and $\mathrm{Sr}^{2+}$ ions (in group V) during mixture analysis because :(1) Sodium ions will react with acid radicals(2) Concentration of $\mathrm{CO}_{3}^{2-}$ ions is very low(3) Mg $^{2+}$ ions will also be precipitated(4) $\mathrm{Na}^{+}$ ions will interfere with the detection of $\mathrm{Ca}^{2+}, \mathrm{Ba}^{2+}, \mathrm{Sr}^{2+}$ ions JEE(Main)-2013 online-P-1

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Sol. (3)According to reactivity series Mg can be displaced by Na.

Q. Which of the following statements is incorrect?(1) $\mathrm{Fe}^{2+}$ ion also gives blood red colour with $\mathrm{SCN}^{-}$ ion(2) Cupric ion reacts with excess of ammonia solution to give deep blue colour of $\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}$ ion.(3) $\mathrm{Fe}^{3+}$ ion gives blood red colour with $\mathrm{SCN}^{-}$ ion.(4) On passing $\mathrm{H}_{2} \mathrm{S}$ into $\mathrm{Na}_{2} \mathrm{ZnO}_{2}$ solution, a white ppt of ZnS is formed.

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Sol. (1)$\mathrm{Fe}^{+2}+\mathrm{SCN}^{-} \longrightarrow$ No colour

Q. Values of dissociation constant, $\mathrm{K}_{\mathrm{a}}$ are given as follows :(1) $\mathrm{NO}_{2}^{-}<\mathrm{CN}^{-}<\mathrm{F}^{-}$(2) $\mathrm{F}^{-}<\mathrm{CN}^{-}<\mathrm{NO}_{2}^{-}$(3) $\mathrm{NO}_{2}^{-}<\mathrm{F}^{-}<\mathrm{CN}^{-}$(4) $\mathrm{F}^{-}<\mathrm{NO}_{2}^{-}<\mathrm{CN}^{-}$ JEE(Main)-2013 online-P-2

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Sol. (4)Order of basic strength according to Ka value.$\mathrm{CN}^{-}>\mathrm{NO}_{2}^{-}>\mathrm{F}^{-}$

Q. Identify incorrect statement(1) Copper (I) compounds are colourless except where colour results from charge transfer(2) Copper (I) compounds are diamagnetic(3) $\mathrm{Cu}_{2} \mathrm{S}$ is black(4) $\mathrm{Cu}_{2} \mathrm{O}$ is colourless JEE(Main)-2013 online-P-3

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Sol. (4)$\mathrm{Cu}_{2} \mathrm{O}$ is red.

Q. Given :$\mathrm{E}_{1 / 2 \mathrm{C}_{2} / \mathrm{Cl}^{-}}^{0}=1.36 \mathrm{V}, \mathrm{E}_{\mathrm{Cr}^{3+} / \mathrm{Gr}}^{0}=-0.74 \mathrm{V} ; \quad \mathrm{E}_{\mathrm{G}_{2} \mathrm{O}_{7}^{2-} / \mathrm{Cr}^{3+}}^{0}=1.33 \mathrm{V}, \mathrm{E}_{\mathrm{MnO}_{4}^{-} / \mathrm{Mn}^{2+}}^{0}=1.51 \mathrm{V}$The correct order of reducing power of the species $\left(\mathrm{Cr}, \mathrm{Cr}^{3+}, \mathrm{Mn}^{2+} \text { and } \mathrm{Cl}^{-}\right)$ will be : JEE(Main)-2013 online-P-3(1) $\mathrm{Mn}^{2+}<\mathrm{Cl}^{-}<\mathrm{Cr}^{3+}<\mathrm{Cr}$(2) $\mathrm{Cr}^{3+}<\mathrm{Cl}^{-}<\mathrm{Mn}^{2+}<\mathrm{Cr}$(3) $\mathrm{Cr}^{3+}<\mathrm{Cl}^{-}<\mathrm{Cr}<\mathrm{Mn}^{2+}$(4) $\mathrm{Mn}^{2+}<\mathrm{Cr}^{3+}<\mathrm{Cl}^{-}<$

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Sol. (1)Order of reducing power acoording to E$\mathrm{Cr}>\mathrm{Cr}^{+3}>\mathrm{Cl}^{-}>\mathrm{Mn}^{+2}$

Q. Which one of the following cannot function as an oxidising agent ?(1) $\mathrm{NO}_{3}^{-}(\mathrm{aq})$( 2) $\mathrm{I}^{-}$(3) $\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}$( 4) $\mathrm{S}_{(\mathrm{S})}$ JEE(Main)-2014 online_P-2

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Sol. (2)$\mathrm{I}^{-}$ has lowest oxidation state.

Q. Which of the following statements about $\mathrm{Na}_{2} \mathrm{O}_{2}$ is not correct ?(1) $\mathrm{Na}_{2} \mathrm{O}_{2}$ oxidises $\mathrm{Cr}^{3+}$ to $\mathrm{CrO}_{4}^{2-}$ in acid medium(2) It is diamagnetic in nature(3) It is the super oxide of sodium(4) It is a derivative of $\mathrm{H}_{2} \mathrm{O}_{2}$ JEE(Main)-2014 online_P-2

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Sol. (2)$\mathrm{Na}_{2} \mathrm{O}_{2}$ is peroxide

Q. Given :-$\mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{e}^{-} \rightarrow \mathrm{Fe}^{2+}(\mathrm{aq}) ; \mathrm{E}^{\circ}=+0.77 \mathrm{V}$$\mathrm{A}^{3+}(\mathrm{aq})+3 \mathrm{e}^{-} \rightarrow \mathrm{Al}(\mathrm{s}) ; \mathrm{E}^{\circ}=-1.66 \mathrm{V}$$\mathrm{Br}_{2}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Br}^{-} ; \mathrm{E}^{\circ}=+1.09 \mathrm{V}$Considering the electrode potentials, which of the following represents the correct order of reducing power?(1) $\mathrm{Al}<\mathrm{Fe}^{2+}<\mathrm{Br}^{-}$(2) $\mathrm{Al}<\mathrm{Br}^{-}<\mathrm{Fe}^{2+}$(3) $\mathrm{Fe}^{2+}<\mathrm{Al}<\mathrm{Br}^{-}$$(4) \mathrm{Br}^{-}<\mathrm{Fe}^{2+}<\mathrm{Al} JEE(Main)-2014 online-P-2 Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (4)Order of reducing power according to \mathrm{E}^{\circ} \mathrm{Br}^{-}<\mathrm{Fe}^{+2}<\mathrm{Al} Q. Consider the following equilibriumWhite precipitate of AgCl appears on adding which of the following?(1) \mathrm{NH}_{3}(2) Aqueous NaCl(3) Aqueous \mathrm{NH}_{4} \mathrm{Cl}(4) AqueousHNO _{3} JEE(Main)-2014 online-P-2 Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (4)\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{4}\right] \mathrm{Cl}+\mathrm{aqH} \mathrm{NO}_{3} \longrightarrow \mathrm{AgCl} \downarrow Q. Copper becomes green when exposed to moist air for a long period. This is due to :-(1) the formation of a layer of cupric oxide on the surface of copper.(2) the formation of basic copper sulphate layer on the surface of the metal(3) the formation of a layer of cupric hydroxide on the surface of copper.(4) the formation of a layer of basic carbonate of copper on the surface of copper. JEE(Main)-2014 online-P-3 Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (4)\mathrm{Cu} \frac{\text { Moist }}{\mathrm{air}}>\mathrm{CuCO}_{3} \cdot \mathrm{Cu}(\mathrm{OH})_{2} Q. The correct statement for the molecule, \mathrm{CsI}_{3}, is:(1) it contains \mathrm{Cs}^{3+} and \mathrm{I}^{-} ions(2) dit contains \mathrm{Cs}^{3+} and \mathrm{I}^{-} ions(3) it is a covalent molecule(4) it contains \mathrm{Cs}^{+} and \mathrm{I}_{3}^{-} ions JEE(Main)-2014 Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (4)\mathrm{CsI}_{3} \longrightarrow \mathrm{Cs}^{+}+\mathrm{I}_{3}^{-} Q. Which one of the following exhibits the largest number of oxidation states ?(1) Mn(25)(2) V(23)(3) Cr (24)(4) Ti (22) JEE(Main)-2014 online-P-3 Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (1)Mn has maximum number of oxidation state. Q. Hydrogen peroxide acts both as an oxidising and as a reducing agent depending upon the nature of the reacting species. In which of the following cases \mathrm{H}_{2} \mathrm{O}_{2} acts as a reducing agent in acid medium ? :- JEE(Main)-2014 online-P-3(1) \mathrm{MnO}_{4}^{-}(2) \mathrm{SO}_{3}^{2-}(3) KI(4) \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (1)\mathrm{H}_{2} \mathrm{O}_{2} acts as a reducing agent in acidic medium in pressure of \mathrm{MnO}_{4}^{-} Q. Consider the reaction\mathrm{H}_{2} \mathrm{SO}_{3(\mathrm{aq})}+\mathrm{Sn}_{(\mathrm{aq})}^{4+}+\mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})} \rightarrow \mathrm{Sn}_{(\mathrm{aq})}^{2+}+\mathrm{HSO}_{4(\mathrm{aq})}^{-}+3 \mathrm{H}_{(\mathrm{aq})}^{+}Which of the following statements is correct?(1) \mathrm{H}_{2} \mathrm{SO}_{3} is the reducing agent because it undergoes oxidation(2) \mathrm{H}_{2} \mathrm{SO}_{3} is the reducing agent because it undergoes reduction(3) \mathrm{Sn}^{4+} is the reducing agent because it undergoes oxidation(4) \mathrm{Sn}^{4+} is the oxidizing agent because it undergoes oxidation JEE(Main)-2014 online-P-4 Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (1) Q. How many electrons are involved in the following redox reaction ?\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+\mathrm{Fe}^{2+}+\mathrm{C}_{2} \mathrm{O}_{4}^{2-} \rightarrow \mathrm{Cr}^{3+}+\mathrm{Fe}^{3+}+\mathrm{CO}_{2} (Unbalanced)(1) 3 (2) 4 (3) 5 (4) 6 JEE(Main)-2014 online-P-4 Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (4)6 e^{-} are involved in this reaction Q. Amongst the following, identify the species with an atom in +6 oxidation state:(1) \left[\mathrm{MnO}_{4}\right]^{-}(2) \left[\mathrm{Cr}(\mathrm{CN})_{6}\right]^{3-}(3) \mathrm{Cr}_{2} \mathrm{O}_{3}(4) \mathrm{Cr} \mathrm{o}_{2} \mathrm{Cl}_{2} JEE(Main)-2014 online-P-4 Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (4)\mathrm{CrO}_{2} \mathrm{Cl}_{2} \mathrm{x}+(-4)+(-2)=0$$\mathrm{x}=+6$

Q. Which of the following salts is the most basic in aqueous solution ?( 1) $\mathrm{CH}_{3} \mathrm{COOK}$( 2) $\mathrm{FeCl}_{3}$(3) $\mathrm{Pb}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2}$(4) $\mathrm{Al}(\mathrm{CN})_{3}$ JEE(Main)-2018

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Sol. (1)

Surface Chemistry – JEE Main Previous Year Questions with Solutions
JEE Main Previous Year Papers Questions of Chemistry with Solutions are available at eSaral. Practicing JEE Main chapter wise questions of Chemistry will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas. Simulator Previous Years AIEEE/JEE Main Questions
Q. Which of the following statements is incorrect regarding physisorptions ?(1) Under high pressure it results into multi molecular layer on adsorbent surface(2) Enthalpy of adsorption $\left(\Delta \mathrm{H}_{\text {adsorption }}\right)$ is low andpositive(3) It occurs because of Van der Waal’s forces(4) More easily liquefiable gases are adsorbed readi AIEEE-2009

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Sol. (2)$\Delta \mathrm{H}$ is negative

Q. According to Freundlich adsorption isotherm, which of the following is correct ?(1) $\frac{\mathrm{x}}{\mathrm{m}} \propto \mathrm{p}^{0}$(2) $\frac{\mathrm{x}}{\mathrm{m}} \propto \mathrm{p}^{1}$(3) $\frac{\mathrm{x}}{\mathrm{m}} \propto \mathrm{p}^{1 / \mathrm{n}}$(4) All the above are correct for different ranges of pressure AIEEE-2012

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Sol. (4)

Q. The coagulating power of electrolytes having ions $\mathrm{Na}^{+}, \mathrm{Al}^{3+}$ and $\mathrm{Ba}^{2+}$ for aresenic sulphide sol increases in the order :-(1) $\mathrm{Al}^{3+}<\mathrm{Ba}^{2+}<\mathrm{Na}^{+}$(2) $\mathrm{Na}^{+}<\mathrm{Ba}^{2+}<\mathrm{Al}^{3+}$.(3) $\mathrm{Ba}^{2+}<\mathrm{Na}^{+}<\mathrm{Al}^{3+}$(4) $\mathrm{Al}^{3+}<\mathrm{Na}^{+}<\mathrm{Ba}^{2+}$ JEE-Main 2013

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Sol. (2)According to hardley schuzle rule

Q. For a linear plot of log(x/m) versus log p in a Freundlich adsorption isotherm, which of the following statements is correct ? (k and n are constants)(1) log (1/n) appears as the intercept(2) Both k and 1/n appear in the slope term(3) 1/n appears as the intercept(4) Only 1/n appears as the slope JEE-Main 2016

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Sol. (4)According to Freundlich isotherm

Q. The Tyndall effect is observed only when following conditions are satisfied :-(a) The diameter of the dispersed particles is much smaller than the wavelength of the lighused.(b) The diameter of the dispersed particle is not much smaller than the wavelength of the lightused.(c) The refractive indices of the dispersed phase and dispersion medium are almost similar inmagnitude.(d) The refractive indices of the dispersed phase and dispersion medium differ greatly inmagnitude.(1) (a) and (d)(2) (b) and (d)(3) (a) and (c)(4) (b) and (c) JEE – Main – 2017

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Sol. (2)As per NCERT book (fact)

Solid State – JEE Main Previous Year Questions with Solutions
JEE Main Previous Year Papers Questions of Chemistry with Solutions are available at eSaral. Practicing JEE Main chapter wise questions of Chemistry will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas. SimulatorPrevious Years AIEEE/JEE Mains Questions
Q. The edge length of a face centered cubic cell of an ionic substance is 508 pm. If the radius of the cation is 110 pm, the radius of the anion is :-(1) 144 pm (2) 288 pm (3) 398 pm (4) 618 pm Aieee-10

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Sol. (1)radius of cation = 110 pmif cation present 2n OV$a=2\left(r_{+}+r_{-}\right)$508 = 2[110 + x]x = 144 pm

Q. Percentages of free space in cubic close packed structure and in body centered packed structure are respectively :-(1) 48% and 26% (2) 30% and 26% (3) 26% and 32% (4) 32% and 48% Aieee-10

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Sol. (3)Theory Based

Q. Copper crystallises in fcc lattice with a unit cell edge of 361 pm. The radius of copper atom is:-(1) 181 pm (2) 108 pm (3) 128 pm (4) 157 pm Aieee-11 / JEE-Main-ONLINE 2013

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Sol. (3)

Q. In a face centred cubic lattice, atom A occupies the corner positions and atom B occupies the face centre positions. If one atom of B is missing from one of the face centred points, the formula of the compound is :-(1) $\mathrm{A}_{2} \mathrm{B}_{3}$( 2) $\mathrm{A}_{2} \mathrm{B}_{5}$(3) $\mathrm{A}_{2} \mathrm{B}$(4) $\mathrm{AB}_{2}$ Aieee-11 / JEE-Main-ONLINE 2014

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Sol. (2)

Q. Lithium forms body centred cubic structure. The length of the side of its unit cell is 351 pm. Atomic radius of the lithium will be :- (1) 152 pm (2) 75 pm (3) 300 pm (4) 240 pm Aieee-12

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Sol. (1)

Q. The radius of a calcium ion is 94 pm and of the oxide ion is 146 pm. The possible crystal structure of calcium oxide will be :-(1) Octahedral        (2) Tetrahedral          (3) Pyramidal           (4) Trigonal Jee-Main (online)-12

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Sol. (1)

Q. Ammonium chloride crystallizes in a body centred cubic lattice with edge length of unit cell of 390 pm. If the size of chloride ion is180 pm, the size of ammonium ion would be:(1) 158 pm (2) 174 pm (3) 142 pm (4) 126 pm Jee-Main (online)-12

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Sol. (1)

Q. A solid has ‘bcc’ structure. If the distance of nearest approach between two atoms is 1.73 Å, the edge length of the cell is :-(1) 314.20 pm (2) 216 pm (3) 200 pm (4) 1.41 pm Jee-Main (online)-12

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Sol. (3)

Q. Among the following the incorrect statement is :- (1) Density of crystals remains unaffecfted due to Frenkel defect(2) In BCC unit cell the void space is 32%(3) Electrical conductivity of semiconductors and metals increases with increase in temperature(4) Density of crystals decreases due to Schottky defect Jee-Main (online)-12

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Sol. (3)Theory Based

Q. In a face centred cubic lattice, atoms of A form the corner points and atoms of B form the face centred points. If two atoms of A are missing from the corner points, the formula of the ionic compound is(1) $\mathrm{AB}_{2}$( 2) $\mathrm{AB}_{3}$(3) $\mathrm{AB}_{4}$(4) $\mathrm{A}_{2} \mathrm{B}_{5}$ Jee-Main (online)-13

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Sol. (3)

Q. Which one of the following statements about packing in solids is incorrect ?(1) Void space in ccp mode of packing is 26%(2) Coordination number in hcp mode of packing is 12(3) Void space in hcp mode of packing is 32%(4) Coordination number in bcc mode of packing is 8 Jee-Main (online)-13

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Sol. (3)Theory based

Q. An element having an atomic radius of 0.14 nm crystallizes in an fcc unit cell. What is the length of a side of the cell ?(1) 0.96 nm (2) 0.4 nm (3) 0.24 nm (4) 0.56 nm Jee-Main (online)-13

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Sol. (2)

Q. Expermentally it was found that a metal oxide has formula $\mathrm{M}_{0.98} \mathrm{O}$. Metal M, is present as $\mathrm{M}^{2+}$ and $\mathbf{M}^{3+}$ in its oxide. Fraction of the metal which exists as $\mathrm{M}^{3+}$ would be :-(1) 7.01%          (2) 4.08%           (3) 6.05%           (4) 5.08 Jee-Main (offline)-13

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Sol. (2)

Q. The total number of octahedral void(s) per atom present in a cubic close packed structure is :-(1) 1          (2) 2              (3) 3              (4) 4 Jee-Main (online)-14

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Sol. (1)Theory based

Q. In a monoclinic unit cell, the relation of sides and angles are respectively(1) $\mathrm{a} \neq \mathrm{b} \neq \mathrm{c}$ and $\alpha \neq \beta \neq \gamma \neq 90^{\circ}$(2) $a \neq b \neq c$ and $\beta=\gamma=90^{\circ} \neq \alpha$(3) $a=b \neq c$ and $\alpha=\beta=\gamma=90^{\circ}$(4) $a \neq b \neq c$ and $\alpha=\beta=\gamma=90^{\circ}$ Jee-Main (online)-14

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Sol. (2)Theory based

Q. The appearance of colour in solid alkali metal halides is generally due to :(1) Frenkel defect (2) F-centres (3) Schottky defect (4) Interstitial position Jee-Main (online)-14

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Sol. (2)Theory based

Q. CsCl crystallises in body centred cubic lattice. if ‘a’ is its edge length then which of the following expression is correct :(1) $\mathrm{r}_{\mathrm{Cs}^{+}}+\mathrm{r}_{\mathrm{Cl}^{-}}=\frac{\sqrt{3}}{2} \mathrm{a}$(2) $\mathrm{r}_{\mathrm{Cs}^{+}}+\mathrm{r}_{\mathrm{Cl}^{-}}=\sqrt{3} \mathrm{a}$(3) $\quad \mathrm{r}_{\mathrm{Cs}^{+}}+\mathrm{r}_{\mathrm{Cl}^{-}}=3 \mathrm{a}$(4) $\mathrm{r}_{\mathrm{Cs}^{+}}+\mathrm{r}_{\mathrm{Cl}^{-}}=\frac{3 \mathrm{a}}{2}$ Jee-Main (offline)-14

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Sol. (1)

Q. A metal crystallises in a face centred cubic structure. If the edge length of its unit cell is ‘a’, the closest approach between two atoms in metallic crystal will be :- (1) 2a(2) $2 \sqrt{2} \mathrm{a}$(3) $\sqrt{2} \mathrm{a}$(4) $\frac{\mathrm{a}}{\sqrt{2}}$ JEE – Main 2017

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Sol. (4)

Q. Which type of ‘defect’ has the presence of cations in the interstitial sites ?(1) Vacancy defect(2) Frenkel defect(3) Metal deficiency defect(4) Schottky defect JEE – Main 2018

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Sol. (2)In Frenkel defect, some of ion (usually cation due to their small size) missing from their correct position and occupies position in interstitial.

Nuclear Chemistry – JEE Main Previous Year Questions with Solutions
JEE Main Previous Year Papers Questions of Chemistry with Solutions are available at eSaral. Practicing JEE Mains chapter wise questions of Chemistry will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.
Electrochemistry – JEE Main Previous Year Questions with Solutions
JEE Main Previous Year Papers Questions of Chemistry with Solutions are available at eSaral. Practicing JEE Main chapter wise questions of Chemistry will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas. Simulator Previous Years AIEEE/JEE Mains Questions
Q. Given :$\mathrm{E}_{\mathrm{Fe}^{-3} / \mathrm{Fe}}^{\circ}=-0.036 \mathrm{V}, \quad \mathrm{E}_{\mathrm{Fe}^{-2} / \mathrm{Fe}}^{\circ}=-0.439 \mathrm{V}$. The value of standard electrode potential for the change.$\mathrm{Fe}^{+3}_{(\mathrm{aq})}+\mathrm{e}^{-} \longrightarrow \mathrm{Fe}^{+2}_{(\mathrm{aq})}$ will be :-(1) 0.770 V (2) –0.27 V (3) –0.072 V (4) 0.385 V AIEEE 2009

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Sol. (1)$\mathrm{E}^{\circ}=\frac{3 \times-0.036+2 \times 0.439}{1}=0.77 \Rightarrow \mathrm{A}$

Q. The Gibbs energy for the decomposition of $\mathrm{Al}_{2} \mathrm{O}_{3}$ at $500^{\circ} \mathrm{C}$ is as follows :$\frac{2}{3} \mathrm{Al}_{2} \mathrm{O}_{3} \longrightarrow \frac{4}{3} \mathrm{Al}+\mathrm{O}_{2}, \mathrm{AG}=+966 \mathrm{KJ} \mathrm{mol}^{-1}$ The potential difference needed for electrolytic reduction of $\mathrm{Al}_{2} \mathrm{O}_{3}$ at $500^{\circ} \mathrm{C}$ is at least :-(1) 5.0 V      (2) 4.5 V       (3) 3.0 V         (4) 2.5 V AIEEE 2010

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Sol. (4)$966 \times 10^{3}=4 \times 96500 \times \mathrm{E} \Rightarrow \mathrm{E}=2.5 \mathrm{V}$

Q. Resistance of 0.2 M solution of an electrolyte is $50 \Omega$. The specific conductance of the solution is 1.3 S $\mathrm{m}^{-1}$. If resistance of the 0.4M solution of the same electrolyte is $260 \Omega$, its molar conductivity is :-(1) $6250 \mathrm{Sm}^{2} \mathrm{mol}^{-1}$(2) $6.25 \times 10^{-4} \mathrm{S} \mathrm{m}^{2} \mathrm{mol}^{-1}$(3) $625 \times 10^{-4} \mathrm{S} \mathrm{m}^{2} \mathrm{mol}^{-1}$(4) $62.5 \mathrm{S} \mathrm{m}^{2} \mathrm{mol}^{-1}$ AIEEE 2011/JEE-MAIN 2014

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Sol. (2)

Q. The reduction potential of hydrogen half-cell will be negative if :-(1) $\mathrm{p}\left(\mathrm{H}_{2}\right)=2 \operatorname{atm}\left[\mathrm{H}^{+}\right]=1.0 \mathrm{M}$(2) $\mathrm{p}\left(\mathrm{H}_{2}\right)=2$ atm and $\left[\mathrm{H}^{+}\right]=2.0 \mathrm{M}$(3) $\mathrm{p}\left(\mathrm{H}_{2}\right)=1$ atm and $\left[\mathrm{H}^{+}\right]=2.0 \mathrm{M}$(4) $\mathrm{p}\left(\mathrm{H}_{2}\right)=1$ atm and $\left[\mathrm{H}^{+}\right]=1.0 \mathrm{M}$ AIEEE 2011

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Sol. (1)

Q. The standard reduction potentials for $\mathrm{Zn}^{2+} / \mathrm{Zn}, \mathrm{Ni}^{2+} / \mathrm{Ni}$ and $\mathrm{Fe}^{2+} / \mathrm{Fe}$ are $-0.76,-0.23$ and – 0.44 V respectively. The reaction $\mathrm{X}+\mathrm{Y}^{+2} \rightarrow \mathrm{X}^{2+}+\mathrm{Y}$ will be spontaneous when(1) X = Zn, Y = Ni(2) X = Ni, Y = Fe(3) X = Ni, Y = Zn(4) X = Fe, Y = Zn

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Sol. (1)For spontaneous reaction $\mathrm{E}^{\circ}>0 \mathrm{so}(\mathrm{A})$

Q. Given :Based on the data given above, strongest oxidising agent will be :(1) Cl–(2) $\mathrm{Cr}^{3+}$(3) $\mathrm{Mn}^{2+}$(4) $\mathrm{MnO}_{4}^{-}$ JEE-Mains 2013

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Sol. (4)Higher the SRP, stronger will be oxidising agentHence, $\mathrm{MnO}_{4}^{-}$ is stronger oxidising agent.

Q. The equivalent conductance of NaCl at concentration C and at infinite dilution are $\lambda_{\mathrm{C}}$ and $\lambda_{\infty}$ , respectively. The correct relationship between $\lambda_{\mathrm{C}}$ and $\lambda_{\infty}$ is given as:(1) $\lambda_{\mathrm{C}}=\lambda_{\infty}-(\mathrm{B}) \sqrt{\mathrm{C}}$(2) $\lambda_{\mathrm{C}}=\lambda_{\infty}+(\mathrm{B}) \sqrt{\mathrm{C}}$(3) $\lambda_{\mathrm{C}}=\lambda_{\infty}+(\mathrm{B}) \mathrm{C}$(4) $\lambda_{\mathrm{C}}=\lambda_{\infty}-(\mathrm{B}) \mathrm{C}$ JEE-Mains 2014

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Sol. (1)Fact

Q. At 298 K, the standard reduction potentials are 1.51 V for $\mathrm{MnO}_{4}-| \mathrm{Mn}^{2+}$ , 1.36 V for $\mathrm{Cl}_{2} | \mathrm{Cl}^{-}$, 1.07 V for $\mathrm{Br}_{2} | \mathrm{Br}^{-},$ and $0.54 \mathrm{V}$ for $\mathrm{I}_{2} | \mathrm{I}^{-} .$ At $\mathrm{pH}=3$, permanganate is expected to oxidize $\left(\frac{\mathrm{RT}}{\mathrm{F}}=0.059 \mathrm{V}\right):-$(1) $\mathrm{Cl}^{-}$ and $\mathrm{Br}^{-}$(2) $\mathrm{Cl}^{-}, \mathrm{Br}^{-}$ and $\mathrm{I}^{-}$(3) $\mathrm{Br}^{-}$ and $\mathrm{I}^{-}$(4) I- only JEE-Mains (online) 2015

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Sol. (3)

Q. A variable, opposite external potential $\left(\mathrm{E}_{\mathrm{ext}}\right)$ is applied to the cell$\mathrm{Zn}\left|\mathrm{Zn}^{2+}(1 \mathrm{M}) \| \mathrm{Cu}^{2+}(1 \mathrm{M})\right| \mathrm{Cu}$, of potential 1.1 V. When $\mathrm{E}_{\mathrm{ext}}<1.1 \mathrm{V}$ and $\mathrm{E}_{\mathrm{ext}}>1.1 \mathrm{V}$ , respectively electrons flow from :(1) anode to cathode in both cases(2) anode to cathode and cathode to anode(3) cathode to anode in both cases(4) cathode to anode and anode to cathode JEE-Mains (online) 2015

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Sol. (2)Fact

Q. Two Faraday of electricity is passed through a solution of $\mathrm{CuSO}_{4}$. The mass of copper deposited at the cathode is :(at. mass of Cu = 63.5 amu)(1) 2g         (2) 127 g          (3) 0 g          (4) 63.5 g JEE-Mains 2015

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Sol. (4)$2 \mathrm{F}=2 \mathrm{eq}=1$ mole $=63.5 \mathrm{gm}$

Q. Galvanization is applying a coating of :-(1)Zn         (2) Pb         (3) Cr        (4) Cu JEE-Mains 2016

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Sol. (1)Galvanization is the process of applying a protective zinc coating of steel or iron, to prevent rusting.

Q. Given(1)Cr(2) $\mathrm{Mn}^{2+}$(3) $\mathrm{Cr}^{3+}$(4) $\mathrm{Cl}^{-}$ JEE-Mains 2017

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Sol. (1)Since $\mathrm{Cr}^{+3}$ is having least reducing potential, so Cr is the best Reducing agent.

Q. How long (approximate) should water be electrolysed by passing through 100amperes current so that the oxygen released can completely burn 27.66 g of diborane ?(Atomic weight of B = 10.8 u)(1)0.8 hours(2) 3.2 hours(3) 1.6 hours(4) 6.4 hours JEE-Mains 2018

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Sol. (2)


text:
# Mutually inaccessible galaxies

It is said the farthest galaxy we can see in one direction is beyond the cosmic horizon of the farthest galaxy we can see in the opposite direction; they can't see each other. Doesn't this contradict the law of addition of velocities? The sum of two velocities cannot be greater than $c$ in our frame, and so the mutual velocity also cannot be greater than $c$, using the addition law again.

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text:
# how to render latex symbol in tex.stackexchange.com? [duplicate]

this piece of latex can be rendered correctly on math.stackexchange.com

$\operatorname{var}[\boldsymbol{\mathrm{AD}}+\mathbf{b}] = \boldsymbol{\mathrm{A}} \operatorname{var}[\boldsymbol{\mathrm{D}}]\boldsymbol{\mathrm{A}}^T$

gives this figure

how to render it correctly on tex.stackexchange.com?

• We don't use MathJax: we're interested in TeX, not in approximate representations of its output. – egreg Jul 2 '19 at 9:52

You cannot render tex code directly. tex.stackexchange has not enabled mathjax

If you want to have an image in your post, compile it on your computer and upload a screen shot


text:
## Calculating Empirical Formula with Given Amounts of Each Element

Wilson 2E
Posts: 100
Joined: Sat Aug 17, 2019 12:18 am

### Calculating Empirical Formula with Given Amounts of Each Element

When calculating the empirical formula of a compound and the masses of each element are given, is it necessary to convert the amount of each into a percentage of the actual mass then use the % out of 100g to find the mol ratio, or can you just go from the amounts given to the number of moles of each without calculating Mass Percentage Composition?

Chem_Mod
Posts: 18880
Joined: Thu Aug 04, 2011 1:53 pm
Has upvoted: 714 times

### Re: Calculating Empirical Formula with Given Amounts of Each Element

Your method is acceptable. Keep in mind that when using mass %, you can easily double check that all of the mass of the sample is accounted for. When you convert mass straight to moles, just make sure that all masses given add up to the total sample mass.

Joseph Saba
Posts: 154
Joined: Thu Jul 11, 2019 12:16 am

### Re: Calculating Empirical Formula with Given Amounts of Each Element

Personally, I like using the 100 g method because if there is such a question that says " there is 40%C, 10%O, 20%H, and an unknown amount of a 4th element, it makes it way easier to go from there inorder to calculate the empirical formula.

KarineKim2L
Posts: 100
Joined: Fri Aug 30, 2019 12:16 am

### Re: Calculating Empirical Formula with Given Amounts of Each Element

No, you do not need to convert the masses of each into percentages when masses are given, unless it asks for mass percent.

abby hyman
Posts: 54
Joined: Thu Jul 25, 2019 12:16 am

### Re: Calculating Empirical Formula with Given Amounts of Each Element

Joseph Saba wrote:Personally, I like using the 100 g method because if there is such a question that says " there is 40%C, 10%O, 20%H, and an unknown amount of a 4th element, it makes it way easier to go from there inorder to calculate the empirical formula.

I agree with this, the 100 g method is the easiest because you are able easily convert the percentages into grams.

Nyari Muchaka_Discussion 4A
Posts: 53
Joined: Thu Sep 19, 2019 12:17 am

### Re: Calculating Empirical Formula with Given Amounts of Each Element

No, you do not have to, the 100g method just makes it easier to check the work and execute the problem.

zfinn
Posts: 106
Joined: Fri Aug 30, 2019 12:16 am

### Re: Calculating Empirical Formula with Given Amounts of Each Element

If the masses are already given I usually just convert to moles and go from there since it eliminates the extra steps of converting into percentages!

Charlotte Coplon 1K
Posts: 5
Joined: Thu Jul 11, 2019 12:15 am

### Re: Calculating Empirical Formula with Given Amounts of Each Element

I don't think that you need to use the 100g method but I agree with the above responses that it simplifies the values in the exercise. Say you had an exercise such as F11, where the amount of Oxygen is not stated. Using the 100g method makes it a lot easier to calculate the missing mass of oxygen. With 100g you get 31.91gK, 28.93gCl and (100-60.84)g O. However, the 100g method comes in handy when the masses are not explicit in the exercise, but you need to find the mass percentage composition.


text:
# naginterfaces.library.sparse.real_symm_basic_setup¶

naginterfaces.library.sparse.real_symm_basic_setup(method, precon, n, tol, maxitn, anorm, sigmax, maxits, monit, sigcmp='N', norm=None, weight='N', iterm=1, sigtol=0.01)[source]

real_symm_basic_setup is a setup function, the first in a suite of three functions for the iterative solution of a symmetric system of simultaneous linear equations. real_symm_basic_setup must be called before the iterative solver, real_symm_basic_solver(). The third function in the suite, real_symm_basic_diag(), can be used to return additional information about the computation.

These three functions are suitable for the solution of large sparse symmetric systems of equations.

For full information please refer to the NAG Library document for f11gd

https://www.nag.com/numeric/nl/nagdoc_28.7/flhtml/f11/f11gdf.html

Parameters
methodstr

The iterative method to be used.

Lanczos method (SYMMLQ).

Minimum residual method (MINRES).

preconstr, length 1

Determines whether preconditioning is used.

No preconditioning.

Preconditioning.

nint

, the order of the matrix .

tolfloat

The tolerance for the termination criterion.

If , is used, where is the machine precision.

Otherwise is used.

maxitnint

The maximum number of iterations.

anormfloat

If , the value of to be used in the termination criterion (2) ().

If , and or , is estimated internally by real_symm_basic_solver().

If , is not referenced.

It has no effect if .

sigmaxfloat

If , the value of .

If , is estimated by real_symm_basic_solver() when either or termination criterion (3) () is employed, though it will be used only in the latter case.

Otherwise, or if , is not referenced.

maxitsint

The maximum iteration number for which is computed by bisection (see also Notes). If or , or if , is not referenced.

Suggested value: when is of the order of its default value .

monitint

If , the frequency at which a monitoring step is executed by real_symm_basic_solver(): the current solution and residual iterates will be returned by real_symm_basic_solver() and a call to real_symm_basic_diag() made possible every iterations, starting from the ()th. Otherwise, no monitoring takes place.

There are some additional computational costs involved in monitoring the solution and residual vectors when the Lanczos method (SYMMLQ) is used.

sigcmpstr, length 1, optional

Determines whether an estimate of , the largest singular value of the preconditioned matrix of the coefficients, is to be computed using the bisection method on the sequence of tridiagonal matrices generated during the iteration. Note that when a preconditioner is not used.

If (see ), i.e., when is supplied, the value of is ignored.

is to be computed using the bisection method.

The bisection method is not used.

If the termination criterion (3) is used, requiring , an inexpensive estimate is computed and used (see Notes).

It is not used if .

normNone or str, length 1, optional

Note: if this argument is None then a default value will be used, determined as follows: if : ; otherwise: .

If or , defines the matrix and vector norm to be used in the termination criteria.

Use the norm.

Use the norm.

Use the norm.

It has no effect if .

weightstr, length 1, optional

Specifies whether a vector of user-supplied weights is to be used in the vector norms used in the computation of termination criterion (2) (): , where , for . The suffix denotes the vector norm used, as specified by the argument . Note that weights cannot be used when , i.e., when criterion (3) is used.

User-supplied weights are to be used and must be supplied on initial entry to real_symm_basic_solver().

All weights are implicitly set equal to one. Weights do not need to be supplied on initial entry to real_symm_basic_solver().

It has no effect if .

itermint, optional

Defines the termination criterion to be used.

Use the termination criterion defined in (2) (both conjugate gradient and Lanczos (SYMMLQ) methods).

Use the termination criterion defined in (3) (Lanczos method (SYMMLQ) only).

It has no effect if .

sigtolfloat, optional

The tolerance used in assessing the convergence of the estimate of when the bisection method is used.

If , the default value is used.

The actual value used is .

If or , is not referenced.

It has no effect if .

Returns
commdict, communication object

Communication structure.

Raises
NagValueError
(errno )

On entry, and .

Constraint: .

(errno )

On entry, , , and .

Constraint: if and , .

(errno )

On entry, , and .

Constraint: if and , .

(errno )

On entry, , and .

Constraint: if and , .

(errno )

On entry, , and .

Constraint: if and , .

(errno )

On entry, .

Constraint: .

(errno )

On entry, .

Constraint: .

(errno )

On entry, .

Constraint: .

(errno )

On entry, and .

Constraint: if , .

(errno )

On entry, and .

Constraint: if , .

(errno )

On entry, and .

Constraint: if , .

(errno )

On entry, .

Constraint: or .

(errno )

On entry, .

Constraint: or .

(errno )

On entry, .

Constraint: , or .

(errno )

On entry, .

Constraint: or .

(errno )

On entry, .

Constraint: or .

(errno )

On entry, .

Constraint: , or .

(errno )

real_symm_basic_setup has been called out of sequence: either real_symm_basic_setup has been called twice or real_symm_basic_solver() has not terminated its current task.

Notes

The suite consisting of the functions real_symm_basic_setup, real_symm_basic_solver() and real_symm_basic_diag() is designed to solve the symmetric system of simultaneous linear equations of order , where is large and the matrix of the coefficients is sparse.

real_symm_basic_setup is a setup function which must be called before real_symm_basic_solver(), the iterative solver. The third function in the suite, real_symm_basic_diag() can be used to return additional information about the computation. One of the following methods can be used:

For this method (see Hestenes and Stiefel (1952), Golub and Van Loan (1996), Barrett et al. (1994) and Dias da Cunha and Hopkins (1994)), the matrix should ideally be positive definite. The application of the Conjugate Gradient method to indefinite matrices may lead to failure or to lack of convergence.

2. Lanczos Method (SYMMLQ)

This method, based upon the algorithm SYMMLQ (see Paige and Saunders (1975) and Barrett et al. (1994)), is suitable for both positive definite and indefinite matrices. It is more robust than the Conjugate Gradient method but less efficient when is positive definite.

3. Minimum Residual Method (MINRES)

This method may be used when the matrix is indefinite. It seeks to reduce the norm of the residual at each iteration and often takes fewer iterations than the other methods. It does however require slightly more memory.

The CG and SYMMLQ methods start from the residual , where is an initial estimate for the solution (often ), and generate an orthogonal basis for the Krylov subspace , for , by means of three-term recurrence relations (see Golub and Van Loan (1996)). A sequence of symmetric tridiagonal matrices is also generated. Here and in the following, the index denotes the iteration count. The resulting symmetric tridiagonal systems of equations are usually more easily solved than the original problem. A sequence of solution iterates is thus generated such that the sequence of the norms of the residuals converges to a required tolerance. Note that, in general, the convergence is not monotonic.

In exact arithmetic, after iterations, this process is equivalent to an orthogonal reduction of to symmetric tridiagonal form, ; the solution would thus achieve exact convergence. In finite-precision arithmetic, cancellation and round-off errors accumulate causing loss of orthogonality. These methods must, therefore, be viewed as genuinely iterative methods, able to converge to a solution within a prescribed tolerance.

The orthogonal basis is not formed explicitly in either method. The basic difference between the Conjugate Gradient and Lanczos methods lies in the method of solution of the resulting symmetric tridiagonal systems of equations: the conjugate gradient method is equivalent to carrying out an (Cholesky) factorization whereas the Lanczos method (SYMMLQ) uses an factorization.

Faster convergence for all the methods can be achieved using a preconditioner (see Golub and Van Loan (1996) and Barrett et al. (1994)). A preconditioner maps the original system of equations onto a different system, say

with, hopefully, better characteristics with respect to its speed of convergence: for example, the condition number of the matrix of the coefficients can be improved or eigenvalues in its spectrum can be made to coalesce. An orthogonal basis for the Krylov subspace , for , is generated and the solution proceeds as outlined above. The algorithms used are such that the solution and residual iterates of the original system are produced, not their preconditioned counterparts. Note that an unsuitable preconditioner or no preconditioning at all may result in a very slow rate, or lack, of convergence. However, preconditioning involves a trade-off between the reduction in the number of iterations required for convergence and the additional computational costs per iteration. Also, setting up a preconditioner may involve non-negligible overheads.

A preconditioner must be symmetric and positive definite, i.e., representable by , where is nonsingular, and such that in (1), where is the identity matrix of order . Also, we can define and . These are formal definitions, used only in the design of the algorithms; in practice, only the means to compute the matrix-vector products and to solve the preconditioning equations are required, that is, explicit information about , or their inverses is not required at any stage.

The first termination criterion

is available for both conjugate gradient and Lanczos (SYMMLQ) methods. In (2), or and denotes a user-specified tolerance subject to , where is the machine precision. Facilities are provided for the estimation of the norm of the matrix of the coefficients , when this is not known in advance, used in (2), by applying Higham’s method (see Higham (1988)). Note that cannot be estimated internally. This criterion uses an error bound derived from backward error analysis to ensure that the computed solution is the exact solution of a problem as close to the original as the termination tolerance requires. Termination criteria employing bounds derived from forward error analysis could be used, but any such criteria would require information about the condition number which is not easily obtainable.

The second termination criterion

is available only for the Lanczos method (SYMMLQ). In (3), is the largest singular value of the (preconditioned) iteration matrix . This termination criterion monitors the progress of the solution of the preconditioned system of equations and is less expensive to apply than criterion (2). When is not supplied, facilities are provided for its estimation by . The interlacing property and Gerschgorin’s theorem provide lower and upper bounds from which can be easily computed by bisection. Alternatively, the less expensive estimate can be used, where by Gerschgorin’s theorem. Note that only order of magnitude estimates are required by the termination criterion.

Termination criterion (2) is the recommended choice, despite its (small) additional costs per iteration when using the Lanczos method (SYMMLQ). Also, if the norm of the initial estimate is much larger than the norm of the solution, that is, if , a dramatic loss of significant digits could result in complete lack of convergence. The use of criterion (2) will enable the detection of such a situation, and the iteration will be restarted at a suitable point. No such restart facilities are provided for criterion (3).

Optionally, a vector of user-specified weights can be used in the computation of the vector norms in termination criterion (2), i.e., , where , for . Note that the use of weights increases the computational costs.

The MINRES algorithm terminates when the norm of the residual of the preconditioned system , , where is the preconditioned matrix.

The termination criteria discussed are not robust in the presence of a non-trivial nullspace of , i.e., when is singular. It is then possible for to grow without limit, spuriously satisfying the termination criterion. If singularity is suspected, more robust functions can be found in submodule opt.

The sequence of calls to the functions comprising the suite is enforced: first, the setup function real_symm_basic_setup must be called, followed by the solver real_symm_basic_solver(). The diagnostic function real_symm_basic_diag() can be called either when real_symm_basic_solver() is carrying out a monitoring step or after real_symm_basic_solver() has completed its tasks. Incorrect sequencing will raise an error condition.

References

Barrett, R, Berry, M, Chan, T F, Demmel, J, Donato, J, Dongarra, J, Eijkhout, V, Pozo, R, Romine, C and Van der Vorst, H, 1994, Templates for the Solution of Linear Systems: Building Blocks for Iterative Methods, SIAM, Philadelphia

Dias da Cunha, R and Hopkins, T, 1994, PIM 1.1 — the parallel iterative method package for systems of linear equations user’s guide — Fortran 77 version, Technical Report, Computing Laboratory, University of Kent at Canterbury, Kent, UK

Golub, G H and Van Loan, C F, 1996, Matrix Computations, (3rd Edition), Johns Hopkins University Press, Baltimore

Hestenes, M and Stiefel, E, 1952, Methods of conjugate gradients for solving linear systems, J. Res. Nat. Bur. Stand. (49), 409–436

Higham, N J, 1988, FORTRAN codes for estimating the one-norm of a real or complex matrix, with applications to condition estimation, ACM Trans. Math. Software (14), 381–396

Paige, C C and Saunders, M A, 1975, Solution of sparse indefinite systems of linear equations, SIAM J. Numer. Anal. (12), 617–629


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# How to use the POSIX standard file descriptors stdin, stdout and stderr

(, en)

This concept is older than I am, from times where things were much simpler. You want to figure out what standard input (stdin), standard output (stdout) and standard error (stderr) are? Let’s start with basic concepts:

1. Most Linux or Unix systems adhere to the POSIX standard.
2. The POSIX standard defines (among other things) three standard file descriptors: stdin, stdout and stderr.

## What is a file desciptor?

File descriptors typically have non-negative integer values, with negative values being reserved to indicate “no value” or error conditions. – Wikipedia

To summarise: if you “open” a file, you get an integer back, a file descriptor. Some default file descriptors are usually available. They serve as communication channels between a process and its environment, aka your console/terminal.

file descriptor name
0 stdin
1 stdout
2 stderr

If you run ls, it prints the output to your terminal, using stdout.

$ls -l total 24 -rw-r--r-- 1 jwb staff 6 5 Nov 00:06 a -rw-r--r-- 1 jwb staff 20 5 Nov 00:06 b -rw-r--r-- 1 jwb staff 37 5 Nov 00:07 c  You can also send data to a process via stdin and some commands send error messages to stderr. ## Redirection You can redirect these “streams” to (and from) files or other processes. In bash it looks like this: # redirect stdin from a file to command; and stdout and stderr of a command to files command <stdin.txt 2>stderr.txt >stdout.txt  You can even redirect one stream into another. Here we redirect stderr to stdout which then gets written to stdout_and_stderr.txt. command >stdout_and_stderr.txt 2>&1  Be aware, the order is important! The command below sends stdout to the file, but stderr is redirected to stdout. command 2>&1 >stdout_only.txt  When is this useful? Nearly never, except you want to mute stdout and continue (aka “pipe”, see further below for explanation) with stderr: command 2>&1 >/dev/null | grep ... # example: ls -z 2>&1 >/dev/null | grep invalid  A detailed explanation you can find in the Redirections chapter of the bash manual. Note that zsh is behaving slightly different. You have to unset the multios option first: # zsh-specific unsetopt multios ls -z 2>&1 >/dev/null | grep invalid  ## Pipelines A pipeline is a sequence of one or more commands separated by one of the control operators | (the pipe symbol) or |&. – bash manual We’ll focus on |. Here we pipe stdout of 1st command to stdin of 2nd command: $ cmd1 | cmd2

How can you visualise it?

Of course you can mix and match redirection and pipe operations.

Have fun!


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# Coordinate vectors for different size matrices

Hi am having trouble with this question

Let $A = \left[\begin{array}{cc}-1&3&-4&-3\\2&-1&0&-3\end{array}\right]$ Find the coordinate vector for the matrix $A \cdot A^{T}$ with respect to the standard basis for $R^{2 \times 2}$

I figured out the $A \cdot A^{T} = \left[\begin{array}{cc}5&-5&-4&-3\\-5&10&12&-6\\-4&12&16&-12\\-3&-6&12&18\end{array}\right]$ and the standard basis for $R^{2 \times 2}$ is $\left[\begin{array}{cc}1&0\\0&0\end{array}\right], \left[\begin{array}{cc}0&1\\0&0\end{array}\right], \left[\begin{array}{cc}0&0\\1&0\end{array}\right], \left[\begin{array}{cc}0&0\\0&1\end{array}\right]$but that is it I get stuck at this point because i cant figure out how $2\times2$ matrices can result to a $4 \times 4$ matrix

-
You have a mistake: you matrix $\;A\;$ is $\;2\times 4\implies A^t\;$ is $\;4\times 2\;$, and thus their product $\;AA^t\;$ must be $\;2\times 2\;$ ... –  DonAntonio Oct 14 '13 at 4:03

$$AA^t=\begin{pmatrix}-1&\;\;3&-4&-3\\\;\;2&-1&\;\;0&-3\end{pmatrix}\begin{pmatrix}-1&\;\;2\\\;\;3&-1\\-4&\;\;0\\-3&-3\end{pmatrix}=\begin{pmatrix}35&4\\4&14\end{pmatrix}$$
THANKS A LOT makes sense now I did $A^{t} \times A$ thinking it would be the same –  MATRIXAREHARD Oct 14 '13 at 4:14
Nop. $\;A^tA\;$ is a $\;4\times 4\;$ matrix...which is what you calculated, of course. The order in matrix product matters a lot. –  DonAntonio Oct 14 '13 at 4:15


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# Is there a name, or any specific jargon for a precipitation equilibrium involving a solid reactant?

Let's say you have some ions in solution, for example Na+(aq) and OH-(aq), and some partially dissolved solid, MgCO3(s), Mg+2(aq) and CO3-2(aq), which precipitates Mg(OH)2(s) until equilibrium.

So you have two solubility equilibriums interacting with each other:

$$\ce{ MgCO_3(s) <=> Mg^{+2}(aq) + CO_3^{-2}(aq) }$$

and

$$\ce{ Mg^{+2}(aq) + 2OH^-(aq) <=> Mg(OH)_2(s) }$$

overall (with all spectator ions removed):

$$\ce{ MgCO_3(s) + 2OH^-(aq) <=> Mg(OH)_2(s) + CO_3^{-2}(aq) }$$

Is there a name for this type of reaction, and what method would I use to calculate the equilibrium point, with what physical properties?

I would call it something like a "binary solubility equilibrium". And I guess if I know the starting OH concentration and the solubility constants of the two solids, I could work out the equilibrium?

Any clues and key words appreciated. Thanks.

AFAIK, such a reaction has no name. However, if one were so inclined, one could call it a replacement reaction, because the $$\ce{MgCO3}$$ is being converted into $$\ce{Mg(OH)2}$$.
To calculate the "equilibrium point," which I take to mean the concentration of all the ions at equilibrium, we would need to know the $$k_\text{sp}$$ of both $$\ce{MgCO3}$$ and $$\ce{Mg(OH)2}$$, as well as the $$\ce{pH}$$ of the solution (which gives us $$[\ce{OH^-}]$$).
The solubility products of $$\ce{MgCO3}$$ and $$\ce{Mg(OH)2}$$ are $$10^{-7.8}$$, which is approximately $$1.58 \cdot 10^{-8}$$, and $$5.61\cdot 10^{-12}$$, respectively.
To calculate how much magnesium hydroxide precipitates, we would first need to find the concentrations of $$[\ce{Mg^{2+}}]$$ and $$[\ce{CO3^{2-}}]$$ using the $$k_\text{sp}$$ of magnesium carbonate. This gives us the solubility of $$\ce{MgCO3}$$, which is $$1.26 \cdot 10^{-4} \,\, \ce{mol lit^{-1}}$$. That means that we have $$[\ce{Mg^{2+}}] = 1.26 \cdot 10^{-4} \,\, \ce{mol lit^{-1}}$$ in the beaker.
Now, to find out how much $$\ce{Mg(OH)2}$$ will precipitate, we need to find the reaction quotient $$Q_c$$ of the reaction. $$Q_c = [\ce{Mg^{2+}}][\ce{OH^-}]^2$$ (I cannot calculate $$Q_c$$ here because it depends on $$[\ce{OH^-}]$$, which will have to be given.)
If $$Q_c > k_\text{sp}$$, then the reaction $$\ce{Mg(OH)_2(s) <=> Mg^{+2}(aq) + 2OH^-(aq)}$$ will proceed in the backward direction and magnesium hydroxide will precipitate out. This will continue until the concentration of magnesium ions decreases to such a value that $$Q_c = k_\text{sp}$$, at which point the reaction will be at equilibrium. Here, I've assumed that the concentration is too large to be appreciably changed by it precipitating out of the solution. The exact value of $$[\ce{Mg^{2+}}]$$ as well as how many moles of $$\ce{Mg(OH)2}$$ will precipitate can be calculated with this assumption. (However, if we want, we can do it without this assumption too. The calculations would become more messy and involve solving a quadratic.)


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