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vidyapeeth-logo                     MOLE CONCEPT PART I


1.              2.               Column – I(A) A gaseous organic compound containing

C = 52.17%, H = 13.04% & 0 = 34.78% (by weight) having molar mass 46 g/mol.

(B) 0.3 g of an organic compound containing

C, H and 0 on combustion yields 0.44 g of CO2 and 0,18 g of FLO. with two 0 atoms per molecule.

(C) A hydrocarbon containing C = 42.857% and H = 57.143% (by mole) containing 3C atoms per molecule.

(D) A hydrocarbon containing 10.5 g carbon per gram of hydrogen having vapour density 46.

Column I

(A) Zn(s) + 2HCl(aq) —› ZnCl2(s) + H2 g)above reaction is carried out by taking 2 moles each of Zn and HCI

(B) AgNO3(aq) + HCI(aq) AgCl(s) + HNO3(g) above reaction is carried out by taking 170 g AgNO3 and 18.25 g HCI (Ag = 108)

(C) CaCO3(s) CaO(s) + CO2(9) 100 g CaCO3 is decomposed

(D) 2KCIO3(s) -+ 2KCI(s) + 302(9) 2/3 moles of KCIO3 decomposed

Column – II

(p)   One mole of compound contains 4N, atoms of Hydrogen.

(q)   The empirical formula of the compound is same as its molecule formula.

(r)     Combustion products of one mole of compound contains larger number of moles of CO2 than that of

(s)    002 gas produced by the combustion of 0.25 mole of compound occupies a volume of 11.2 L at NTP.

Column II

(p) 50% of excess reagent left

(q) 22.4 L of gas at STP is liberated

(r)  1 moles of solid (product) obtained.

(s) HCI is the limiting reagent



Column-I         Column-II

  • Molarity          (p) Dependent on temperature
    MA x nA
  • Molality (q)m   m             x 100

-A…A       ..B…B

  • Mole fraction            (r) Independent of temperature


(ID) Mass %                                                                                        (s) x____ B mx 1000

Where MA , MB are molar masses, nA , nB are no of moles & XA X, are mole fractions of solute and solvent respectively.

4.                                                                                                                          Column-I(A)      100 ml of 0.2 M AICI3 solution + 400 ml of 0.1 M HCI solution

(B)      50 ml of 0.4 M KCI + 50 ml H2O

(C)     30 ml of 0.2 M K2SO4 + 70 ml H2O

(D)     200 ml 24.5% (w/v) H2SO4

Column-II(p)            Total concentration of cation(s) = 0.12 M

(q)            [SO42-] = 0.06 M

(r)             [SO42-] = 2.5 M

(s)            [01 ] = 0.2 M



Read the following comprehension carefully and answer the questions.

Comprehension # 1

A chemist decided to determine the molecular formula of an unknown compound. He collects following informations :

  • Compounds contains 2 : 1 ‘H’ to ‘0’ atoms(number of atoms).
  • Compounds has 40% C by mass

(111) Approximate molecular mass of the compound is 178 g

(IV) Compound contains C, H and 0 only.

  1. What is the % by mass of oxygen in the compound

(A) 53.33%                           (B) 88.88%                          (C) 33.33%                          (ID) None of these

  1. What is the empirical formula of the compound

(A) CH30                       (B) CH20                      (C) C2H20                     (D) CH302

  1. Which of the following could be molecular formula of compound

(A) C61-1606                            (B) C6H1206                             (C) C6H14012                                (ID) C51-11406

Comprehension # 2

According to the Avogadro’s law, equal number of moles of gases occupy the same volume at identical condition of temperature and pressure. Even if we have a mixture of non-reacting gases then Avogadro’s law is still obeyed by assuming mixture as a new gas.

Now let us assume air to consist of 80% by volume of Nitrogen (N2) and 20% by volume of oxygen (02). If air is taken at STP then its 1 mol would occupy 22.4 L. 1 mol of air would contain 0.8 mol of N2 and 0.2 mol of 02 hence the mole fractions of N2 and 02 are given by XN2 = 0.8 , X02 = 0.2

  1. Volume occupied by air at NTP containing exactly 11.2 gm of Nitrogen :

(A) 22.4 L                              (B) 8.96 L                              (C) 11.2 L                            (D) 2.24 L

  1. If air is treated as a solution of 02 and N2 then % W/W of oxygen is :

10                                200                              700                              350

(A) —9                                               (B) 9                                       (C) 9                                          (D) 9

  1. Density of air at NTP is :

9                                             2

(A) 1 g/L                                 (B) 7 g/L                                (C) —7 g/L                               (D) can’t be determined

Comprehension # 3

The concentrations of solutions can be expressed in number of ways; viz : mass fraction of solute (or mass percent), Molar concentration (Molarity) and Molal concentration (molality). These terms are known as concentration terms and also they are related with each other i.e. knowing one concentration term for the solution, we can find other concentration terms also. The definition of different concentration terms are given below

Molarity : It is number of moles of solute present in one litre of the solution.

Molality : It is the number of moles of solute present in one kg of the solvent

moles of solute

Mole Fraction = moles of solute – moles of solvent

If molality of the solution is given as ‘a’ then mole fraction of the solute can be calculated by


Mole Fraction = a a” MsoNer:  – (a x Msolvent 1000)
1000a +




where a = molality and              = Molar mass of solvent

We can change : Mole fraction H Molality E Molarity

  1. 60 gm of solution containing 40% by mass of NaCI are mixed with 100 gm of a solution containing 15% by
    mass NaCI. Determine the mass percent of sodium chloride in the final solution.

(A) 24.4%                            (B) 78%                                  (C) 48.8%                          (D) 19.68%

  1. What is the molality of the above solution.

(A) 4.4 m                             (B) 5.5 m                      (C) 24.4 m                         (D) none

  1. What is the molarity of solution if density of solution is 1.6 gm/ml

(A) 5.5 M                              (B) 6.67 Nil                             (C) 2.59 M                 (D) none

Comprehension # 4

In chemistry, oxidation and reduction are taken as two mutually exclusive events. For example, if life is oxidation then death is taken as reduction. taking off a flight is oxidation then standing would be reduction and so many other. In brief it is used as redox in chemical science.

There are so many conceptual facts regarding redox such as adding oxygen or oxygenation, removing hydrogen or dehydrogenation, removing electron or dielectronation are fixed for oxidation and their corresponding antonyms would be reduction processes. Simple way of judging whether a monatomic species has under gone oxidation or reduction is to note if the charge number of species has changed. It is possible to assign to an atom in polyatomic species an operative charge number called their oxidation number or state. (0. N. or 0. S.). There is no standard symbol for this quantity so we say it is 1;). An 0. N. is assigned to an element in a compound by assuming that it is present as ion with a characteristic charge for instance oxygen is present as 0(-11) and fluorine as F(-1) and some time it may be hypothetical also. For example


In continuation to our study, species promoting oxidation are named as oxidant and those promoting reduction are termed as reductant. At the same time their equivalent weights is the ratio of their molecular weight and change is 0. N. (AO) involving one molecule/formula unit of the reactant i.e., molecular weight divided by number of electrons lost or gained by one molecule/formula during their respective action.

Based on the above discussion answer the following objective question having one best answer.

  1. Which corresponds to oxidation action

(A) = 0                         (B)              = 0                            (C)              > 0                  (D) AO) < 0

  1. A compound contain P(II), Q(V) R(-11). The possible formula of the compound is

(A) PQR2                       (B) Q2(PR3)2                           (C) P3[QR4J2                          (D) P3(Q4R)2

  1. A compound has 0 number of carbon, 0 number of hydrogen and NJ number of oxygen their equation of finc
    oxidation number (x) of carbon will be
(A)                                                          + 4×02 + =0            (B) x0 + —     =0 2kv (C) @X +                       —                                 = 0 (D) none of these

x                        3




* Marked Questions may have more than one correct option.

1.   How many moles of electron weigh one kilogram :

(A) 6.023 x 1023 1 6.023_______  x      31

(B) 9.108       (C) 9.108 x 1054


1 x 108

(D) 9.108 x 6.023


2.   Which has maximum number of atoms :

(A) 24 g of C (12)                  (B) 56 g of Fe (56)               (C) 27 g of Al (27)                 (D) 108 g Ag (108)

3.   Amongst the following, the pair having both the metals in their highest oxidation state is :

(A) [Fe(CN)6]3- and [Co(CN)6]3-                                                (B) CrO2C12 and Mn04

(C) TiO2 and Mn02                                                                  (D) [IVInC14]2- and [NiF6]2-

4.    Paragraph for Question Nos. (i) to (iii)

Chemical reactions involve interaction of atoms and molecules. A large number of atoms/molecules (approximately 6.023 x 1023) are present in a few grams of any chemical compound varying with their atomic molecular masses. To handle such large numbers conveniently, the mole concept was introduced. This concept has implications in diverse areas such as analytical chemistry, biochemistry, electrochemistry and radiochemistry. The following example illustrates a typical case, involving chemical / electrochemical reaction, which requires a clear understanding of the mole concept.

A 4.0 molar aqueous solution of NaCI is prepared and 500 mL of this solution is electrolysed. This leads to the evolution of chlorine gas at one of the electrodes (atomic mass : Na = 23, Hg = 200 ; 1 Faraday = 96500 coulombs).

**[At the anode :                  2CI > C12 + 2e –

At the cathode :                   Na’ + e —> Na

Na + Hg NaHg (sodium amalgam)] ** (These reactions were not present in IIT-JEE paper)

  • The total number of moles of chlorine gas evolved is :

(A) 0.5                                    (B) 1.0                                   (C) 2.0                                 (D) 3.0

  • If the cathode is a Hg electrode, the maximum weight (g) of amalgam formed from this solution is :

(A) 200                                         (B) 225                                  (C) 400                      (D) 446

  • The total charge (coulombs) required for complete electrolysis is :

(A) 24125                              (B) 48250                             (C) 96500                  (D)193000

5.   Given that the abundances of isotopes S Fe, “Fe and 57Fe are 5%, 90% and 5%, respectively, the atomic mass

of Fe is :

(A) 55.85                               (B) 55.95                              (C) 55.75                     (D) 56.05

6.   A student performs a titration with different burettes and finds titre values of 25.2 mL, 25.25 mL, and 25.0 mL.

7.   Among the following, the number of elements showing only one non-zero oxidation state is :                              0,     CI,         F,                                    N, P,                                         Sn,       Ti,                    Na,       Ti

8.   Dissolving 120 g of urea (mol. wt. 60) in 1000 g of water gave a solution of density 1.15 g/mL. The molarity of

the solution is :

(A) 1.78 M                    (B) 2.00 M                    (C) 2.05 M                   (D) 2.22 M

9.   2% (w/w) HCI stock solution has a density of 1.25 g mL-1. The molecular weight of HCI is 36.5 g mol-1. The volume (mL) of stock solution required to prepare a 200 mL solution of 0.4 M HCI is :


  1. Number of atoms in 560g of Fe (atomic mass 56g mol-1) is :

(1) Twice that of 70g N (2) Half that of 20g H    (3) Both (1) and (2)      (4) None of these

  1. In an organic compound of molar mass 108 g mol-1 C,H and N atoms are present in 9 : 1 : 3.5 by weight.

Molecular formula can be :

(1) C6H8N2                               (2) C71-1,DN                           (3) C5H6N3                               (4) C4H18N3

  1. When KMnO4 acts as an oxidising agent and ultimately forms Mn042-, Mn02, Mn203 and Mn2+, then the

number of electrons transferred in each case is :

(1)4, 3, 1, 5                  (2) 1, 5. 3, 7                 (3) 1, 3, 4, 5                (4) 3, 5, 7, 1

4.         Which of the following is a redox reaction?(1) NaCI + KNO3 ____ > NaNO3 + KCI

(2) CaC2O4 + 2 HCI ____ > CaCl2 H2C204

(3) Mg (OH)2 + 2NH4CI             MgCl2 + 2NH4OH

(4) Zn + 2 AgCN __  2 Ag + Zn (CN)2


  1. Which of the following concentration factor is affected by change in temperature ?

(1) Molarity                   (2) Molality                (3) Mole fraction(4) Weight fraction

  1. What volume of hydrogen gas at 273 K and 1 atm pressure will be consumed in obtaining 21.6 gm of elemental

boron (atomic mass = 10.8) from the reduction of boron trichioride by hydrogen-

(1) 44.8 lit.                    (2) 22.4 lit.                   (3) 89.6 lit.                   (4) 67.2 lit.

  1. 02 x 1020 molecules of urea are present in 100 ml of its solution. The concentration of urea solution is            (1) 0.001 M               (2) 0.01 M                (3) 0.02 M              (4) 0.1 M
  2. If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass L-

mass of one mole of a substance will

(1) decrease twice                                         (2) increase two fold

(3) remain unchanged                               (4) be a function of the molecular mass of the sL:E.7-E,-

  1. The oxidation state of Cr in [Cr(NH3)4C12]± is :

(1) + 3                          (2) + 2                         (3) + 1                         (4) 0

  1. The oxidation state of chromium in the final product formed by the reaction between KI and acidic _ –

dichromate solution is :

(1) + 4                          (2) + 6                         (3) + 2                         (4) + 3



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