Chemistry by Inam Jazbi: XI MCQs on Mole Concept with Explanatory Answers

XI MCQs on Mole Concept with Explanatory Answers

MCQs on Unit # 1… Mole Concept Set 1

1. Which of the following contains the least numbers of molecules?

(a) 1 g of H₂

(b) 4 g of O₂

(d) 3 g of Cl₂

Explanation; (Answer; d)

Substance containing least number of moles would have least number of molecules. Here 3g of Cl₂ contains least number of moles (0.042), so it has least number of molecules.

2. 1 gram molecule of any gas at STP occupies the volume of

(a) 22.4 cm³

(b) 0.0224 dm³

(d) 100 cm³

Explanation; (Answer; c)

1 gram molecules constitutes one mole of that substance. 1 mole of any gas at STP occupies the fixed volume known as molar gas volume which is equal to 22.4 dm³ or 22400 cm³ or 0.0224 m³.

3. The number of moles of carbon in 200 g of CaCO₃ is

(a) 1 mole

(b) 2 mole

(d) 4 mole

Explanation; (Answer; b)

100 g of CaCO₃ contains 1 mole of C

200 g of CaCO₃ contains 1/100 x 200 = 2 mole of C

4. The volume of semi-mole of a gas at STP is

(a) 11.2 cm³

(b) 11200 cm³

(d) 0.0224 m³

Explanation; (Answer; b)

Semi mole means half mole. One mole of any gas STP occupies 22.4 dm³. Hence half mole of any gas occupies half of 22.4 dm³ i.e. 11.2 dm³. But this is not given in the options. 11.2 dm³ equals to 11200 cm³. So correct option is b.

5. At NTP, 5.6 L of gas weighs 8 grams. The vapour density of a gas is

(a) 32

(b) 40

(d) 8

Explanation; (Answer; c)

We know that NTP is Normal Temperature and Pressure which is defined as a temperature of 200C and 1 atmosphere. Vapor density is defined as the density of a vapor in relation to that of hydrogen. Vapor density can tell us whether a gas is denser or less dense than air.

6. The vapour densities of two gases are in the ratio of 1:3. Their molecular masses are in the ratio of:

(a) 1:3

(b) 1:2

(d) 3:1

Explanation; (Answer; a)

7. The number of atoms of Cr and O are 4.8 x 10¹⁰and 9.6 x 10¹⁰respectively. Its empirical formula is

(a) Cr₂O₃

(b) CrO₂

(d) CrO₅

Explanation; (Answer; b)

Alternative method

Molar ratio or atomic ratio of Cr to O = 4.8 x 10¹⁰ : 9.6 x 10¹⁰ = 1 : 2

Hence, empirical formula = CrO₂

8. A hydrocarbon contains 80% C. The vapour density of compound is 30. The hydrocarbon is

(a) CH₃

(b) C₂H₆

(d) C₄H₈

Explanation; (Answer; a)

Molecular mass = vapour density x 2 = 30 x 2 = 60

Integer (n) = molecular mass/empirical formula mass = 60/15 = 4

Molecular formula = (EF)_n = (CH₃)₄= C₄H₁₂

9. Two elements X (atomic weight = 75) and Y (atomic weight = 16) combine to give a compound having 75.8% of X. The empirical formula of the compounds is

(a) Xy

(b) X₂Y

(d) X₂Y₃

Explanation; (Answer; d)

10. If a compound on analysis was found to contain C = 18.5%, H = 1.55%, Cl = 55.04% and O = 25.81%, then its empirical formula is

(a) CHClO

(b) CH₂ClO

(d) CHCl₂O

Explanation; (Answer; a)

11. An unknown gas has a density 1.96 gL⁻¹. Then gas is

(a) CO

(b) NO₂

(d) SO₂

Explanation; (Answer; c)

Volume of 1 mole of gas at STP = 22.4 L

The molecular mass of CO₂ is 12 + 32 = 44 amu.

12. 2.2 g of an unknown gas is present in a container of 1.12 L. Then unknown gas is

(a) CO₂

(b) CO

(d) Both a and c

Explanation; (Answer; d)

The molecular mass of CO₂ (12 + 32) and N₂O (28+16) = 44 amu. So gas unknown gas may be CO₂ or N₂O.

13. 16 g of oxygen has same number of molecules as in

(a) 16 g of CO

(b) 28 g of N₂

(d) 2 g of He

Explanation; (Answer; b)

According to Avogadro’s law, equal number of moles contains equal number of molecules. 16 g oxygen constitutes its one mole. 28 g of N₂ also constitutes its one mole. So 16 g oxygen (1 mole) has the same number of molecules as in 28 g of N₂ (1 mole).

14. The volume of 3.01 x 10²³ molecules of N₂ gas at STP will be

(a) 3 dm³

(b) 11.2 cm³

(d) 11200 cm³

Explanation; (Answer; d)

15. The mass of one mole of electrons is:

(a) 1.008 mg
(b) 0.55 mg
(c) 0.184 mg
(d) 1.673 mg

Explanation; (Answer; b)

Mass of 1 electron = 9.1×10⁻³¹ kg = 9.1 × 10⁻²⁵ mg

1 mole of electron = 6.02 × 10²³ electrons

Mass of 1 mole of electrons = mass of 1 electron × number of one mole of electron
= (9.1 × 10⁻²⁵) × (6.02×10²³)

= 0.54782mg ≈ 0.55mg

The mass of one mole of electrons = 0.55 mg

16. The number of moles of CO₂which contain 8.0 g of oxygen:

(a) 0.25

(b) 0.50

(d) 1.50

Explanation; (Answer; a)

Mass of 1 mole or Molar mass of CO₂ = 12 x 16 x 2 = 44 g/mol

Mass of oxygen in 32 g of CO₂ = 16 x 2 = 32 g

32 g of oxygen is contained in 1 mole of CO₂

1 g of oxygen is contained in 1/32 mole of CO₂

8 g of oxygen is contained in = 1/32 x 8 = 0.25 mole of CO₂

17. The largest number of molecules are present in:

(a) 3.6 g of water

(b) 4.8 g of C₂H₅OH

(d) 5.4 g of N₂O₅

Explanation; (Answer; a)

The largest number of molecules will be present in that sample which has greatest number of moles.

Hence 3.6 g water having greatest number of moles out of given options will contain largest number of molecules.

18. The formula of KAl(SO₄)₂ represents a total of:

(a)6 atoms

(b)7 atoms

(d)14 atoms

Explanation; (Answer; c)

KAl(SO₄)₂ = 1K + 2Al + 2S + (4 x 2)O= 12 atoms

19. Which will weigh more?

(a) 0.1 mole of sucrose

(b) 1 mole of acetic acid

(d) All have same

Explanation; (Answer; c)

Mass of 0.1 mole of sucrose (C₁₂H₂₂O₁₁) = nM = 0.1 x 342 = 34.2 g

Mass of 0.1 mole of acetic acid (CH₃COOH) = molar mass = 60 g

Mass of molecules of SO₂ = N_p/N_A x M = 6.01x10²³ /6.02x10²³ x 64 = 64 g

20. The number of atoms in 2.4g magnesium and 1.0 g of hydrogen are.

(a) Equal atoms

(b) Different atoms

(d) None of these

Explanation; (Answer; b)

Same number of moles contains same number of particles. 2.4g magnesium constitutes its 0.1 mole and 1.0 g of hydrogen constitutes its 1 mole. Since moles are different, so both of them contains different number of atoms. 1 g H has more atoms than 2.4 g Mg.

21. Objects of the size of an atom can be observed in

(a) an electron microscope

(b) atomic absorption spectrum

(d) visible spectrum

Explanation; (Answer; a)

Objects of the size of an atom can be observed in an electron microscope

22. Two different hydrocarbon each contain the same percentage by mass of hydrogen. It follows that they have the same

(a) Empirical formula

(b) number of atoms in a molecule

(d) relative molecular mass

Explanation; (Answer; a)

Same percentage composition means same empirical formula.

23. % age of oxygen in combustion analysis is calculated by the formula

a) (%age of C + % of H) – 100

(b) (%age of C + % of H) +100

(d) 100 – (%age of C + % of H)

Explanation; (Answer; d)

% age of oxygen in combustion analysis is calculated by the difference method.

24. Which one of the following has same number of electrons?

(a) F^-, Na⁺, Ne, O^2-, N^3-, Mg²⁺

(b) Ca²⁺, Mg²⁺, Al³⁺, Sc³⁺, Ar, Ti⁴⁺

(d) All of them

Explanation; (Answer; a)

Number of electrons in cation = Z – Number of positive charge

Number of electrons in anion = Z + Number of negative charge

25. Molecular ions are produced in mass spectrometer. Which type of molecular ions is more abundant?

(a) Positively charged

(b) Negatively charged

(d) Molecules cannot form ions.

Explanation; (Answer; a)

When the vaporised organic sample passes into the ionization chamber of a mass spectrometer, it is bombarded by a stream of electrons. These electrons have a high enough energy to knock an electron off an organic molecule to form a positive ion. This ion is called the molecular ion symbolized as M⁺. The molecular ions tend to be unstable and some of them break into smaller fragments. All mass spectrometers separate ions based on their mass to charge ratio, and molecules can be ionized and analyzed just like atoms are.

26. Using the Periodic Table for the relative atomic masses, which has the least mass?

(a) 0.1 moles of silicon dioxide, SiO₂

(b) Hemi moles of lithium, Li

(d) 1.0 moles of ammonia, NH₃

Explanation; (Answer; b)

Mass of 0.1 mole of SiO₂ = 0.1 x 60 = 6 g

Mass of 0.5 (hemi) mole of Li = 0.5 x 7 = 3.5 g

Mass of 0.5 (semi) mole of O₂ = 0.5 x 32 = 16 g

Mass of 1.0 mole of NH₃ = 1.0 x 17 = 17 g

27. Which is not represented by 1mole of Nitrogen gas?

(a) 6.023 x 10²³ molecules of N₂

(b) 6.023 x10²³ atoms of N₂

(d) 28 g of N₂

Explanation; (Answer; b)

One mole of N₂ = molar mass = N_A molecules = 2 N_A atoms

One mole of N₂ = 28 g = 6.023 x10²³ molecules = 2 x 6.023 x10²³ (12.046 x 10²³) atoms

28. Out of 1 g of oxygen gas, 1 g of oxygen atoms and 1 g of ozone, maximum number of atoms are present in:

(a) 1 g of oxygen gas

(b) 1 g of ozone
(c) 1 g of oxygen atoms

(d) All have equal number of atoms

Explanation; (Answer; c)

Greater the n, greater is the number of particles. 1 g of O atoms has largest n, so it contains maximum number of atoms.

29. The number of atoms contained 11.2 litre of SO₂ at STP are

(a) 3/2 x 6.023 x 10²³

(b) 2 x 6.023 x 10²³

(d) 4 x 6.023 x 10²³

Explanation; (Answer; a)

30. 2 g of oxygen contains number of atoms equal to that in

(a) 0.5 g of hydrogen gas

(b) 2.3 g of sodium

(d) 4 g of sulphur

Explanation; (Answer; d)

2 g of oxygen constitutes its 0.125 mole (2/16)

0.125 mole of oxygen contains number of atoms equal to 0.125 x 6.02 x 10²³ = 7.525 x 10²²atoms

0.5 g of hydrogen constitutes its 0.25 mole (0.5/2)

2.3 g of sodium constitutes its 0.1 mole (2.3/23)

7 g of nitrogen constitutes its 0.5 mole (7/14)

4 g of sulphur constitutes its 0.125 mole (4/32)

4 g of sulphur and 2 g of oxygen constitutes same number of moles (i.e. 0.125 mole), therefore, they contain same number of atoms.

31. The largest number of molecules is in

(a) 34 g of water

(b) 28 g of CO₂

(d) 54 of N₂O₅

Explanation; (Answer; a)

Greater the number of moles, greater is the number of particles. 34 g of water contains largest mole (1.88), so it contains the largest number of molecules.

(a) 34 g of water constitutes its 1.88 mole

(b) 28 g of CO₂ constitutes its 1 mole

(d) 54 of N₂O₅ constitutes its 0.458 mole

32. 1 mol of CH₄ contains

(a) 6.02 × 10²³ atoms of H

(b) 3.0 g of carbon

(d) 1.81 × 10²³ molecules of CH₄

Explanation; (Answer; c)

We know that 1 mole of any substance has Avogadro number of units in it, like 1 mole of oxygen has Avogadro number of molecules in it.

1 mole of CH₄ contains 1 mole of carbon (1 gram atom of carbon)

1 mole of CH₄ contains 4 moles of hydrogen (4 gram atoms of hydrogen)

1 mole of CH₄ contains 6.023 x 10²³ atoms of carbon

1 mole of CH₄ contains 6.023 x 10²³ x 4 atoms of hydrogen (4N_A atoms of H)

1 mole of CH₄ contains 6.023 x 10²³ x 5 atoms of C and H (5N_A atoms of H)

1 mole of CH₄ contains 6.023 x 10²³ molecules of CH₄

1 mole of CH₄ contains 12 g of carbon

1 mole of CH₄ contains 4 g of hydrogen

33. Which of the following has smallest number of molecules?

(a) 0.1 mol of CO₂ gas

(b) 22 g of CO₂gas

(d) 22.4 mL of CO₂ gas

Explanation; (Answer; a)

Smaller number of moles contains least number of molecules. Since CO₂ has least number of mole out of given options, so it contains least number of molecules.

(a) 0.1 mol of CO₂ gas

(b) 22 g of CO₂gas constitutes its 0.5 mole

(d) 22.4 mL of CO₂ gas constitutes its 1 mole

34. Number of atoms in 558.5 gram Fe (atomic weight of Fe = 55.85 g mol^–1) is

(a) 558.5 × 6.023 × 10²³

(b) half that in 8 g He

(d) 6.023 × 10²²

Explanation; (Answer; c)

Number of moles in 558.5 g of Fe = 558.5/55.85 = 10 mol = 10 N_A atoms

Number of moles in 60 g of C = 60/12 = 5 mol = 5 N_A atoms

35. Which of the following has least mass?

(a) 2 gram atom of nitrogen

(b) 3 x 10²³ atoms of carbon

(d) 5.6 liter of oxygen gas at STP

Explanation; (Answer; b)

(a) 2 gram atom of nitrogen weighs 28 g (14 x 2)

(b) 3 x 10²³ atoms of carbon weighs 6 g (3 x 10²³ /6.02 x 10²³ x 12)

(d) 5.6 liter of oxygen gas at STP weighs 8 g.

36. One mole of any substance contains 6.022 x 10²³atoms/molecules. Number of molecules of H₂SO₄ present in 100 mL of 0.02 M H₂SO₄ solution is ----------------

(a) 12.044 x 10²⁰ molecules

(b) 6.022 x 10²³ molecules

(d) 12.044 x 10²³ molecules

Explanation; (Answer; a)

37. 32 g of oxygen, 28g nitrogen occupy separately the volume of 22.414 dm³

(a) Size of molecules of two gases are same
(b) Masses of molecules two gases are same
(c) Number of atoms of two gases are same
(d) All of these

Explanation; (Answer; c)

32 g O₂ and 28 g N₂constitutes 1 mole of respective gases which contain same of molecules and also atoms (due to same atomicity).

38. Which pair of molecules contains the same number of molecules at STP?

(a) 280 cm³ of CO₂ and 280 cm³ of N₂O

(b) 11.2 dm³of O₂ and 32 g of O₂

(d) 1 gram molecule of N₂ and 5.6 dm³ of Cl₂

Explanation; (Answer; a)

Equal volumes of all gases have equal number of moles and number of molecules.

39. Which of the following has the same number of molecules at STP?

(a) 1 dm³ of N₂ and O₂

(b) 500 cm³ of Cl₂ and O₂

(d) All of them

Explanation; (Answer; a)

Equal volumes of all gases have equal number of moles and number of molecules.

40. A limiting reactant is one which according to the stoichiometric equation?

(a) has excess mass

(b) has least mass

(d) has least number of moles

Explanation; (Answer; d)

A limiting reactant is one which according to the stoichiometric equation has least number of moles

41. In the determination of atomic ratio of the elements, the mole ratios are divided by

(a) Least value of gram atoms of elements

(b) Atomic masses of elements

(d) molecular mass of the compound

Explanation; (Answer; a)

Atomic ratio of elements is the ratio of moles of each element by the smallest number of moles (i.e. least value of gram atoms) giving relative ratio of atoms in the compound

42. Relative atomic mass, Ar

Explanation; (Answer; a)

Relative atomic mass, Ar is the ratio of average mass of an atom of an alement to the 1/12^th mass of one atom of C – 12.

43. Which of the following has maximum mass?

(a) 0.1 gram atom of carbon

(b) Hemi mole of ammonia

(d) 1120 cc of carbon dioxide at STP

Explanation; (Answer; b)

(a) 0.1 gram atom of carbon weights 1.2 g

(b) Hemi mole of ammonia weighs 8.5 g

(d) 1120 cc of carbon dioxide at STP 2.2 g

44. 0.1 mole of Na₃PO₄ completely dissociates in water to produce Na⁺

(a) 6.02 x 10²²

(b) 6.02 x 10²³

(d) 1.806 x 10²²

Explanation; (Answer; c)

1 mole of Na₃PO₄ gives total 4 moles of ions i.e. 3 moles of Na⁺ and 1 mole of PO₄⁻ion.

0.1 mole of Na₃PO₄ gives 3 x 0.1 x N_A Na⁺ ions = 0.3 x 6.02 x 10²³ Na⁺ ions = 1.806 x 10²³ Na⁺ ions

45. The largest number of H⁺ are produced by complete ionization of

(a) 0.1002 moles of HCl

(b) 0.051 moles of H₂SO₄

(d) All of the above

Explanation; (Answer; d)

HCl being a Monoprotic acid gives 1 mole of H⁺ ion on ionization. Hence 0.102 moles of HCl gives 0.102 moles of H⁺ ions.

H₂SO₄ being a diprotic acid gives 2 mole of H⁺ ion on ionization. Hence 0.051 moles of H₂SO₄ gives 0.051 x 2 = 0.102 moles of H⁺ ions.

H₃PO₄ being a triprotic acid gives 3 mole of H⁺ ion on ionization. Hence 0.0334 moles of H₃PO₄ gives 0.0334 x 2 = 0.1002 moles of H⁺ ions.

46. 16 g of S₈ contains:

(a) 6.023 x 10²³ atoms of S

(b) 6.023 x 10²³/8 atoms of S

(d) 6.023 x 10²³x 0.0625 molecules of S

Explanation; (Answer; d)

We know that 1 mole of any substance has Avogadro number of units in it, like 1 mole of oxygen has Avogadro number of molecules in it. Sulphur has an atomic weight of approximately 32 g/mol, so the molar mass of S₈ would be 256 g/mol.

Now, 16 g of S₈ would thus correspond to 0.0625 (16/256) moles and thus the number of molecules in 16 g of S₈ would be 0.0625 times Avogadro’s number i.e. 0.0625 x 6.022 x 10²³ molecules of S₈.

47. Which of the following has the smallest number of molecules

(a) 0.1 moles of CO₂

(b) 2 g of H₂ at STP

(d) 3.4 g of NH₃

Explanation; (Answer; a)

Greater is the number of moles, greater is the number of particles and vice versa.

0.1 moles of CO₂

2 g of H₂ at STP constitutes its one mole

16 g of O₂ gas constitutes its half mole

3.4 g of NH₃ gas constitutes its 0.2 mole

Since 0.1 moles of CO₂ has least number of moles, so it contains the smallest number of molecules.

48. Which of the following weighs the least?

(a) 2.0 gram mole of CO₂

(b) Semi mole of sucrose (C₁₂H₂₂O₁₁)

(d) 1.5 mole of water

Explanation; (Answer; c)

(a) 2.0 gram mole of CO₂ weighs 88 g (44 x 2)

(b) Semi mole of sucrose (C₁₂H₂₂O₁₁) weighs 171 g (342 x 0.5)

(d) 1.5 mole of water weighs 27 g (18 x 1.5)

49.The largest number of molecules is present in

(a) 59 g of nitrogen peroxide

(b) 54 g of water

(d) 58 g of ethyl alcohol

Explanation; (Answer; c)
Greater is the number of moles, greater is the number of particles and vice versa.
(a) 59 g of nitrogen peroxide constitutes its 0.5 mole

(b) 54 g of water constitutes its 3 mole
(c) 56 g of carbon dioxide constitutes its 1.27 mole
(d) 58 g of ethyl alcohol constitutes its 1.26 mole

50. 1g-atom of nitrogen represents

(a) 6.02 × 10²³ N₂ molecules

(b) 11.2 L of N₂ at N.T.P.

(d) 28 g of nitrogen

Explanation; (Answer; b)

Different equivalent of 1 g-atom of Nare

14 g N₂

N_A atoms of N

11.2 liter or dm³

Gram atom signifies gram atomic mass. Gram atomic mass of N is 14 g.

At STP 1 mole of nitrogen gas (N₂) occupies 22.4 liters of N₂ gas

So 14 g of nitrogen occupies 11.2 liters of N₂ gas.

51. Number of atoms in 560 g of Fe (atomic mass 56 g/mol) is

(a) is twice that of 70 g of N

(b) is half as 2 grams of hydrogen

(d) None of these

Explanation; (Answer; a)

Number of atoms in 560 g of Fe = mass/molar mass = 560/56=10 mole

Number of atoms in 70 g of N = mass/molar mass = 79/14 =5 mole

52.Compared to 88g of carbon dioxide, 88g of propane contains

(a) The same number of atoms

(b) The same number of molecules

(d) More molecules

Explanation; (Answer; b)

Both propane and carbon dioxide have same molar mass of 44 g/mol. For same masses, compounds with same molar mass have same number of moles and consequently same number of molecules.

53. The incorrect statement for 14 g of CO is

(a) It occupies 2.24 liter at STP

(b) It corresponds to 0.5 mol of CO

(d) It corresponds to 3.01 x 10²³

Explanation; (Answer; a)

14 g of CO occupies 11.2 liter at STP

14 g of CO corresponds to 0.5 mol of CO

14 g of CO corresponds to same mole of CO and N₂

14 g of CO corresponds to 3.01 x 10²³

54. 10 moles of H₂O contains

(a) 100 moles of bonds

(b) 25 moles of hydrogen bonds

(d) 100 moles of electrons

Explanation; (Answer; d)

10 moles of H₂O contains 20 moles of bonds (2 x 10)

10 moles of H₂O contains 40 moles of hydrogen bonds (4 x 10)

10 moles of H₂O contains 30 atoms (3 x 10)

10 moles of H₂O contains 100 moles of electrons (10 x 10)

55. The maximum number of molecules is present in

(a) 15 L of H₂ gas at STP

(b) 1.5 g of H₂gas

(d) 5 g of O₂ gas

Explanation; (Answer; b)

(a) 15 L of H2 gas at STP constitutes its 0.669 mole

(b) 1.5 g of H2 gas constitutes its 0.669 mole 0.75 mole

(d) 5 g of O₂ gas constitutes its 0.156 mole

Greater number of moles corresponds to greater number of molecules (also atoms for same atomicity). Here 1.5 g of H2 gas has greatest number of moles and hence it contains greatest number of molecules.

56. Which among the following is the heaviest?

(a) one mole of oxygen

(b) 100 amu of uranium

(d) 44 g carbon dioxide

Explanation; (Answer; d)

(a) one mole of oxygen weighs 32 g

(b) 100 amu of uranium weighs 3.95 x 10-20 g

(d) 44 g carbon dioxide

57. If proton number of two atoms is same then it can be concluded that

(a) They are isotopes

(b) Their compounds will be similar in reactivity towards other compounds

(d) Both have same melting point

Explanation; (Answer; b)

Isotopes have same chemical properties

58. A mole of any substance is related to

(a) Number of particles

(b) Mass of a substance

(d) All of these

Explanation; (Answer; d)

A mole of any substance is related to all of these

59. This one of the following pairs has the same number of molecules

(a) 10 g H₂ and 10 g CH₄

(b) 10 g H₂ and 50 g CH₄

(d) 10 g H₂ and 80 g CH₄

Explanation; (Answer; c)

Same number of moles contains same number of molecules.

(a) 10 g H₂ and 10 g CH₄ constitutes 5 mole H₂ and 0.625 mole CH₄

(b) 10 g H₂ and 50 g CH₄ constitutes 5 mole H₂ and 3.125 mole CH₄

(d) 10 g H₂ and 80 g CH₄ constitutes 5 mole H₂ and 5 mole CH₄

60. How many moles of chlorine atoms are present in 0.99 g of C₂H₄Cl₂?

(a) 0.01

(b) 0.0178

(d) 0.75

Explanation; (Answer; c)

61. One mole of carbon-12 has a mass of:

(a) 0.012 kg

(b) 0.0224 kg

(d) 1 kg

Explanation; (Answer; a)

Mass of one mole of carbon-12 is 12 g or 12/1000 = 0.012 kg

62. How many moles are present in 52 g of aspartame (C₁₄H₁₈N₂O₅)?

(a) 0.177

(b) 0.36

(d) 0.54

Explanation; (Answer; a)

Molar mass of aspartame (C₁₄H₁₈N₂O₅) = (12 x 14) + (18x1) + (14x2) + (16x5) = 294 gmol⁻¹

No. of moles (n) = m/M = 52/294 = 0.177 mole

63. The number of moles of CO₂ which contain 16 g of oxygen

(a) 0.50

(b) 1.0

(d) 1.50

Explanation; (Answer; a)

1 mole (44g) of CO₂ contains 12 g carbon (1mole) and 32 g (2mole) oxygen

32 g oxygen is present in 1 mole of CO₂

16 g oxygen is present in =16/32 mole of CO₂ = 0.50 mole

64. A vessel contains 10 g of N₂, 10 g of H₂ and 10 g O₂. Which one has the largest number of atoms?

(a) H₂

(b) N₂

(d) All of these

Explanation; (Answer; a)

Molar masses of N₂, h₂ and O₂ are 28, 2 and 32 respectively. Greater the molar mass lesser is the number of moles and lesser is the number of molecules and number of atoms (for same atomicity). Hence the number of their moles are in their order H₂> N₂> O₂ in their equal masses. The same order follows for number of molecules and number of atoms.

65. What is total volume of 0.5 g of H₂, 16 g of O₂ and 7.0 g of N₂ present in a mixture at STP?

(a) 2.4 dm³

(b) 0.224 dm³

(d) 11.2 dm³

Explanation; (Answer; c)

Species ……………. H₂ : O₂ : N₂

Given masses ……… 0.5 16 7.0

Molar masses ……… 2 32 28

No of moles ………… 2/2 16/32 7.0/28

Total moles …………. 0.25 + 0.5 + 0.25 = 1 mole

Total volume ……….. 1 mole = 22.4 dm³

66. The mass of one mole of electrons in mg is:

(a) 1.008 mg

(b) 1.673 mg

(d) 0.184 mg

Explanation; (Answer; c)

Mass of 1 electron = 9.1 X 10⁻³¹ kg.

1 mole = 6.023 X 10²³

So 1 mole of electrons has a mass of = 9.1 x 10⁻³¹ x 6.023 x 10²³ ≈ 5.481 x 10⁻⁷ kg ≈ 5.481 x 10⁻⁴ g ≈ 5.481 x 10⁻¹ mg ≈ 0.5481 or 0.55 mg

67. The mass of one mole of electrons is:

(a) 1.008 mg

(b) 0.184 mg

(d) 0.55 mg

Explanation; (Answer; d)

Mass of one electron is 9.11 x 10⁻³¹ kg

Mass of one mole (6.02 x 10²³) electron is = 9.11 x 0⁻³¹ x 6.02 x 10²³ = 54.84 x 10⁻⁸ kg or 5.5 x 10⁻⁷ kg

Mass of one mole of electron in mg = 5.5 x 10⁻⁷ kg x 1 x 10⁶ = 5.5 x 10⁻¹ mg or 0.55 mg

68.The number of moles of CO₂which contain 8.0 g of oxygen:

(a) 0.25

(b) 0.50

(d) 1.50

Explanation; (Answer; a)

32 g O = 1 mole of CO₂

8.0 g O = 1/32 x 8.0 = 0.25 mole of CO₂

69. One mole of hydrogen gas at s.t.p. contains how many atoms?

(a) 6.02 x 10²³

(b) 3.01 x 10²⁴

(d) 12.04 x 10⁴⁶

Explanation; (Answer; c)

No of atoms = atomicity x n x N_A = 2 x 1 x N_A = 2N_A = 2 x 6.02 x 10²³ =1.2 x 10²⁴

70. The mass of a single hydrogen atom is:

(a) 1.67x10⁻²⁰g

(b) 1.67x10⁻²⁴g

(d) 1.67x10⁻²³ g

Explanation; (Answer; b)

Mass of single atom = atomic mass/N_A = 1/6.02 x 10²³ = 1.66 x 10⁻²⁴ g

71. 1.12 dm³ of nitrogen gas at STP weighs:

(a) 28 g

(b) 2.8

(d) 14

Explanation; (Answer; c)

22.4 dm³ of nitrogen gas at STP weighs 28 g

11.2 dm³ of nitrogen gas at STP weighs 11.2/22.4 x 28 g = 1.4 g

72. The mass of one atom of coal is:

(a) 1.99 x10⁻²²g

(b) 1.99 x10⁻²³g

(d) 1.99 x10⁻²⁰ g

Explanation; (Answer; b)

Mass of single atom = atomic mass/N_A = 12/6.02 x 10²³ = 1.99 x 10⁻²³ g

73.The weight of one mole of KAl(SO₄)₂.12H₂O is:

(a) 574g g

(b) 584 g

(d) 684 g

Explanation; (Answer; c)

The weight of one mole of KAl(SO₄)₂.12H₂O is equal to its gram formula mass as it is an ionic compound.

KAl(SO₄)₂.12H₂O = 39 + 27 + 2(32) + 8(16) + 12(18) = 474 g

74. The density of oxygen gas at STP is:

(a) 1.43 g/dm³

(b) 2.43 g/dm³

(d) 0.42 g/dm³

Explanation; (Answer; a)

Density of gas = molar mass/molar volume = 32/22.4 = 1.43 gdm-3

75. What is the mass of one mole of bromine molecule?

(a) 80 g

(b) 160 g

(d) 70 g

Explanation; (Answer; b)

Molecular mass of bromine (Br₂) = 80 x 2 = 160 g

76. The volume occupied by a hemi (0.1) mole of a gas at stp is:

(a) 22400 cm³

(b) 5.6 dm³

(d) 22.4 dm³

Explanation; (Answer; c)

Volume of gas at STP = n x molar volume = 0.1 x 22.4 = 2.24 dm³ or 2240 cm³ or 0.00224 m³

77. The number of protons in one molecule of HNO₃ are:

(a) 7

(b)8

(d)32

Explanation; (Answer; c)

No of protons in HNO₃ = 1 + 7 + 3(8) = 32

78. One cm³ oxygen gas at stp contains about:

(a) 1 x 10²⁰atoms

(b) 6 x 10²³atoms

(d) 0.53 x 10¹⁹atoms

Explanation; (Answer; d)

No. of atoms = atomicity x (V_g/V_M) x N_A = 2 (1/22400) N_A = 8.93 x 10^-5 N_A = 8.93 x 10^-5 x 6 x 10²³= 5.37 x 10¹⁸0.537 x 10¹⁹

79. The relative atomic mass of chlorine (Cl) is 35.5 amu, the mass in gram of 0.5 moles of chlorine gas is:

(a) 71 g

(b) 35.5 g

(d) 17.75 g

Explanation; (Answer; b)

Mass = nM = 0.5 x 71 = 35.5 g

80. If 28 g of N₂gas comprises of z molecules; how many molecules are there in 16 g of oxygen?

(a) z

(b) 2 z

(d) ¼ z

Explanation; (Answer; c)

28 g of N₂ gas constitutes its one mole which contains N_A (here z) molecules. 16 g of O₂ constitutes its half mole which contains ½ N_A molecules (i.e. ½ z).

81. What is the relative molecular mass, M_r, of CuSO₄.5H₂O?

(a) 127

(b) 160

(d) 249.5

Explanation; (Answer; d)

relative molecular mass = Sum of atomic masses of all atoms

relative molecular mass CuSO₄.5H₂O = 63.5 + 32 + (16 x 4) + 5(18) = 249.5 amu

82. The total number of ions in FeCl₃ is

(a) 6.02 x 10²³

(b) 12.04 x 10²³

(d) 24.08 x 10²³

Explanation; (Answer; d)

total number of ions in 1 mole = total ions x n x N_A = 4 x 1x N_A= 4N_A = 4 x 6.02 x 10²³ = 24.08 x 10²³or 2.408 x 10²⁴

83. Which of the following has maximum number of atoms?

(a) 1.8 g of H₂O

(b) 8.8 g of CO₂

(d) 12 g of CH₄

Explanation; (Answer; d)

No of atoms from mass = mass/molar x atomicity x N_A

No of atoms 1.8 g of H₂O = 1.8/18 x 2 x N_A = 0.2 N_A

No of atoms 8.8 g of CO₂ = 8.8/44 x 3 x N_A = 0.6 N_A

No of atoms 16 g of O₂ = 16/32 x 2 x N_A = 1 N_A

No of atoms 12 g of CH₄= 12/16 x 5 x N_A = 3.75 N_A

84. The number of moles of CO₂ which contains 8.0 g of oxygen?

(a) 0.25

(b) 0.50

(d) 1.50

Explanation; (Answer; a)

32 g of oxygen is present in 1 mole of CO₂

1 g of oxygen is present in 1/32 mole of CO₂

8 g oxygen is present in 1/32 x 8 mole of CO₂ = 0.25 mole of CO₂

85. An atom has 8 protons and 10 neutrons, the atom is:

(a) N

(b) B

(d) Ne

Explanation; (Answer; c)

Atomic number characterized an element. O has an atomic number of 8.

86.The number of protons in one molecule of H₂SO₄ is:

(a) 40

(b) 50

(d) 30

Explanation; (Answer; b)

Total number of protons = Sum of number of protons of all atoms

Total number of protons in H₂SO₄ = (1 x 2) + 16 + (4 x 8) = 50

87. ----------- is a generic name for hydrogen cations; protons, deuterons and tritons.

(a) Nucleons

(b) Hydride

(d) Hydron

Explanation; (Answer; d)

The generic name for three types of hydrogen cations i.e. protons, deuterons and tritons is Hydron.

88. The number of atoms present in 0.5 moles of Nitrogen atoms is same as in:

(a) 12 g of C

(b) 32 g of S

(d) 8 g of O

Explanation; (Answer; d)

8 of O constitutes its half mole.

89. How many moles of helium gas occupy 22.4 litre at 0ºC and 1 atm pressure?

(a) 0.11

(b) 1.0

(d) 1.11

Explanation; (Answer; b)

At STP (0ºC temperature and 1 atm pressure), one mole of all gases occupy a constant volume of 22.4 litre which is called molar gas volume.

90. The total number of protons in 10 g of calcium carbonate is

(a) 1.5057 x10²⁴

(b) 2.0478 x 10²⁴

(d) 4.0956 x10²⁴

Explanation; (Answer; c)

Protons in 1 mole of CaCO₃ = Z of Ca + Z of C + Z of O x 3 Þ 20 + 6+24 = 50 moles of protons

100 g of CaCO₃ contains 50 moles of protons

1 g of CaCO₃ contains 50/100 moles of protons

10 of CaCO₃ contains = 50/100 x 10 x 6.02 x 10²³ moles of protons = 3.0115 x 10²⁴ protons

91. The formula of hydrogen phosphate of certain metal is MHPO₄. The formula of metal sulphate would be

(a) MSO₄ (b) M₂SO₄ (c) M₃SO₄ (d) M₂(SO₄)₃

Explanation; (Answer; a)

The net charge on HPO₄ ion is 2-. Thus in MPHO₄, the charge on M would be 2+. The charge of SO₄ is 2-, so formula of metal sulphate (M) would be MSO₄.

92. If N_A is Avogadro’s number, then number of valence electrons in 4.2 g of nitride ion (N³⁻) is

(a) 2.4

(b) 4.2

(d) 3.2

Explanation; (Answer; a)

N has 5 valence electrons. It gains 3 electrons to form nitride ion (N³⁻) which now contains total 8 valence electrons.

14 g of nitride ion contains valence electrons = 8 N_A

1 g of nitride ion contains valence electrons = 8N_A/14

4.2 g of nitride ion contains valence electrons = (8N_A/14) x 4.2 = 2.4 N_A

93. Vapour density of a gas is 22. What is its molecular mass?

(a) 33

(b) 44

(d) 11

Explanation; (Answer; b)

Molar mass = 2 x vapour density = 2 x 22 = 44

94. The numerical value of ‘N/n’ (where N is the number of molecules in a given sample of gas and ‘n’ is the number of moles of the gas) is

(a) 8.314

(b) 0.0821

(d) 6.02×10²³

Explanation; (Answer; c)

6.023 ×10²³ molecules (N) = 1 mole (n)

∴ N/n = 1/6.023 × 10²³= 1.62×10⁻²⁴

95. The number of sulphur atoms present in 0.2 moles of S₈ molecules is

(a) 9.63 × 10²²

(b) 4.82 × 10²³

(d) 9.63 × 10²³

Explanation; (Answer; d)

No of atoms = atomicity x n x N_A= 8 x 0.2 x N_A= 1.6 N_A = 1.6 × 6.02 × 10²³= 9.632 × 10²³

96. What is the number of atoms in 0.1 mol of a tetra-atomic gas? (N_A= 6.02 x 10²³ molecules)

(a) 6.022 x 10²²

(b) 6.022 x 10²³

(d) 2.4088 × 10²³

Explanation; (Answer; d)

One mole of any substance contains 6.022 x 10²³ particles (Avogadro's number).

Thus

Number of molecules = number of moles x Avogadro's number

Number of molecules= 0.1 x 6.022 x 10²³= 6.022 x 10²²

Each molecule of tetra-atomic gas contains 4 atoms (atomicity = 4), therefore,

Number of atoms in = atomicity x N_A = 4 x 6.022 x 10²²= 24.088 × 10²² OR 2.4088 × 10²³

97. 1 atom of an element weighs 1.792 × 10⁻²² g. The Atomic mass of the element is

(a) 17.92

(b) 1.192

(d) 64

Explanation; (Answer; c)

1 atom of an element weighs 1.792 × 10⁻²² g

6.02 × 10²³ atom of an element weighs = 1.792 × 10⁻²²× 6.02 × 10²³= 107.878 ≈ 108 g

98. A sample of phosphorus trichloride (PCl₃) contains 1.4 moles of the substance. How many atoms are there in the sample?

(a) 4

(b) 5.6

(d) 3.36 x 10²⁴

Explanation; (Answer; d)

One mole of any substance contains 6.022 x 10²³ particles (Avogadro's number)

Thus

Number of molecules = number of moles x Avogadro's number

Number of molecules = 1.4 × 6.022 × 10²³= 8.431 x 10²³

Each molecule of PCl₃ contains 4 atoms; 1 P atom and 3 Cl atoms, therefore,

Number of atoms in 1.4 mole PCl₃ = 8.431 × 10²³ × 4 = 3.36 × 10²⁴

99. The mass of 1 atom of hydrogen is

(a) 0.5 g

(b) 1.6 × 10⁻²⁴ g

(d) 3.2 × 10⁻²⁴ g

Explanation; (Answer; b)

Mass of 1 atom = atomic mass/N_A = 1/6.02 × 10²³= 1.6 × 10⁻²⁴ g

100. Two containers of the same size are filled separately with H₂ gas and CO₂ gas. Both the containers under the same T and P will contain the same

(a) Number of electrons

(b) Number of molecules

(d) Number of atoms

Explanation; (Answer; b)

Since volume of container is same, so both gases have same volume and same no of moles and consequently same number of molecules.

101. The number of atoms in 4.25 g of NH₃ is approximately

(a) 1 × 10²³

(b) 2 × 10²³

(d) 6 × 10²³

Explanation; (Answer; d)

Number of atoms in a molecule = atomicity x (mass/M) x N_A = 4 x (4.25/17) x N_A = 1N_A = 6 × 10²³

102. One mole of any covalent compounds contains same number of

(a) Atoms

(b) Molecules

(d) Electrons

Explanation; (Answer; b)

covalent compounds consist of discrete molecules.

103. One atom of an element X weighs 6.664 ×10^–23 g. The number of gram atoms in 40 kg of it are

(a) 10

(b) 10000

(d) 1000

Explanation; (Answer; d)

Gram atomic mass = mass of 1 mole atoms × mass of one atom

= (6.022×10²³) × (6.664 ×10⁻²³) = 40.01 g

Gram atoms = moles = mass/molar mass = 40000/40 = 1000

∴ The number of gram atoms in 40 kg = 1000

104. Which one of the following is not an iso-electronic pair?

(a) Ca⁺, K⁺

(b) Na⁺, Ne

(d) O²⁻, Al³⁺

Explanation; (Answer; a)

Iso-electric ions have same number of electrons.

Number of electrons in Ca⁺ = 20 – 1 = 19

Number of electrons in K⁺= 19 – 1 = 18

105. The mass of CO containing the same amount of oxygen as in 88 g of CO₂ is

(a) 56 g

(b) 28 g

(d) 14 g

Explanation; (Answer; c)

Molar mass of CO2 = 44 g

Mass of oxygen in 44 g of CO₂ = 32 g

Mass of oxygen in 88 g of CO₂ = 64 g

Molar mass of CO = 28 g

Mass of O in CO = 16 g

16 g of O = 30 g CO

64 g of O = (28/16) x 64 = 112 g

106. The mass of CO containing the same number of oxygen atoms as are present in 88 g of CO₂ is

(a) 56 g

(b) 28 g

(d) 14 g

Explanation; (Answer; c)

No. of molecule = (m/M) N_A = (88/44) x 6.02 x 10²³ = 1.2046 x 10²⁴ molecules

No of atoms of O = atomicity x no of molecules = 2 x 1.2046 x 10²⁴ = 2.4092 x 10²⁴ atoms

6.022 x 10²³ atoms of O = 28g of CO

Mass = N_p/N_A x M = 2.4092 x 10²⁴ /6.022 x 10²³ x 28 = 112 g

Alternate Method

Molecular mass of CO₂ = 12 + 2 x 16 = 44

1mol of CO₂ = 44 g

44 g of CO₂ contain = 6.022 x 10²³ molecules

88 g of CO₂ contain = 6.022x10²³x 88/44 = 12.046 x 10²³ molecules

Each molecule of CO₂ contain 2 oxygen atoms

No. of oxygen atoms in 12.046 x 10²³ molecules of CO₂ = 2 x 12.046 x 10²³ = 2.4092 x 10²⁴

Now, we have to calculate the mass of CO containing 2.4092 x 10²⁴atoms of oxygen.

Mass of CO containing 6.022 x 10²³ oxygen atoms = 28 g.

Mass of CO containing 2.4092 x 10²³ oxygen atoms = 28/6.022 x 10²³ x 2.4092 x 10²⁴ = 112 g

107. Number of O₂ molecules present in one litre flask at pressure 7.6 × 10^–10 mm of Hg at 0ºC are

(a) 2.69 × 10¹⁰

(b) 2.69 × 10¹¹

(d) None of these

Explanation; (Answer; a)

PV = nRT Þ PV = N_p/N_A RT Þ N_p = PVN_A/RT

N_p = (7.6 × 10^–10 /760 atm) (1 liter) (6.02 x 10²³ molecules-mol^-1)/(0.0821atm-liter-mol^-1 K^-1 )x (0+273 K)

N_p = 1 x 10^-12 x 6.02 x 10²³/5.733

N_p = 2.69 × 10¹⁰ molecules

108. Volume occupied by one molecule of water (density = 1 g cm^–3) is

(a) 9.0 × 10^–23cm³

(b) 6.023 × 10^–23 cm³

(d) 5.5 × 10^–23 cm³

Explanation; (Answer; c)

Density of water (ρ) = 1

Molar mass of water (M) = 18 g mol^–1

Density = Molar mass/molar volume Þ Molar volume = molar mass/density = 18/1 = 18 cm3 mol^–1

6.02 x 10²³ molecules of water = 18 cm³

1 molecules of water = 18/6.02 x 10²³ = 2.989 x 10^–23 or 3 x 10^–23 cm³

109. The diameter of atoms is of the order of

(a) 2 x 10⁻⁵ m

(b) 2 x 10⁻¹⁰ m

(d) 2 x 10⁻³ m

Explanation; (Answer; b)

The atomic size or atomic radius is of the order of 10−7cm or 10−9m or 1 nanometer (nm).

r_atom ≈10⁻¹⁰m
r_nucleus ≈10⁻¹⁵ m

Under most definitions the radii of isolated neutral atoms range between 30 and 300 pm (trillionths of a meter), or between 0.3 and 3 ångströms. Therefore, the radius of an atom is more than 10,000 times the radius of its nucleus (1–10 fm), and less than 1/1000 of the wavelength of visible light (400–700 nm).

110.One mole of ethane and one mole of ethanol have an equal

(a) Number of molecules

(b) Number of electrons

(d) Masses

Explanation; (Answer; c)

One mole of different compounds would have same number of molecules (and also same number of atoms for same atomicity).

111. How many atoms of uranium (U) are present in 1 nanogram (ng) of uranium?

(a) 2.5 x 10²⁰

(b) 5.0 x 10¹⁰

(d) 5.0 x 10³⁰

Explanation; (Answer; c)

1 nanogram (ng) of uranium = 10⁻⁹ g

No. of atoms = n x N_A = (m/M) x N_A = (10⁻⁹/238)x 6.02 x10²³ = 2.5 x 10¹² atoms of uranium

112. How many moles of electrons are found in 401.4 g of P³⁻?

(a) 194

(b) 402

(d) 232.2

Explanation; (Answer; d)

1 mole of P³⁻contains 30.97 g of P³⁻. Atomic number is equal to number of electrons.

The electronic configuration of phosphorus P (Z=15) is 1s² 2s² 2p⁶,3s² 3p³ showing that it has 15 electrons. Phosphorus, P gains three electrons and converts into P³⁻giving total 18 electrons.

No of moles in in 401.4 g of P³⁻= 401.4/30.97 = 12.96 mole

1 mole of P³⁻ ion contains 18 electrons

12.96 mole of P³⁻ ion contains 18 x 12.96 electrons = 233.2 mole of electrons

113.Which one of the following will have largest number of atoms?

(a) 1 g Au_(s)

(b) 1 g Na_(s)

(d) 1 g of Cl_2(g)

Explanation; (Answer; c)

114. 6.3 g of HNO₃ has mass of nitrate ions

(a) 6.2 g

(b) 62 g

(d) 3.1g

Explanation; (Answer; a)

Molar mass of HNO₃ = 63 g

Molar mass of nitrate (NO₃⁻) ion = 62 g

63 g HNO₃ contains 62 g nitrate ion

6.3 g HNO₃ contains (62/63) x 6.3 nitrate ion = 6.2 g nitrate ions

115. 9 g of ice has number of covalent bonds

(a) 6.02 x 10²³

(b) 3.01 x 10²³

(d) 1.505 x 10²³

Explanation; (Answer; a)

Molar mass of ice (H₂O) = 18 g

18 g ice contains = 2 mole covalent bonds (H–O–H)

9 g ice contains (2/18) x 9 mole covalent bonds = 1 mole of covalent bond = 6.02 x 10²³ covalent bonds

116. A well-known ideal gas is enclosed in a container having volume 5603 cm³ at STP. Its mass comes out to be 16 g. The unknown gas is

(a) O₂

(b) CH₄

(d) CO₂

Explanation; (Answer; c)

5603 cm³ of gas at sTP weighs 16 g

22400 cm³ of gas at STP weighs (16/5603)x 22400 (molar mass) = 63.96 ≈ 64 g

The molar mas of SO₂is 64.

117. Which of the following has the same atomic and molecular mass?

(a) bromine

(b) oxygen

(d) nitrogen

Explanation; (Answer; )

118. The wavelength of visible light is 500 nm. In SI unit this value is

(a) 500 x 10⁻⁹ m

(b) 5 x 10⁻⁹ m

(d) 5 x 10⁻⁸ m

Explanation; (Answer; a)

1nm = 10⁻⁹ m

500 nm = 500 x10⁻⁹ m = 5 x 10² x 10⁻⁹ = 5 x 10⁻⁷ m

119. Which of the following contains the least numbers of molecules?

(a) 1 g of H₂

(b) 4 g of O₂

(d) 3 g of Cl₂

Explanation; (Answer; d)

Substance containing least number of moles would have least number of molecules. Here 3g of Cl₂ contains least number of moles (0.042), so it has least number of molecules.

120. 1 gram molecule of any gas at STP occupies the volume of

(a) 22.4 cm³

(b) 0.0224 dm³

(d) 100 cm³

Explanation; (Answer; c)

1 gram molecules constitutes one mole of that substance. 1 mole of any gas at STP occupies the fixed volume known as molar gas volume which is equal to 22.4 dm³ or 22400 cm³ or 0.0224 m³.

121. The number of moles of carbon in 200 g of CaCO₃ is

(a) 1 mole

(b) 2 mole

(d) 4 mole

Explanation; (Answer; b)

100 g of CaCO₃ contains 1 mole of C

200 g of CaCO₃ contains 1/100 x 200 = 2 mole of C

122.The volume of semi-mole of a gas at STP is

(a) 11.2 cm³

(b) 11200 cm³

(d) 0.0224 m³

Explanation; (Answer; b)

123. At NTP, 5.6 L of gas weighs 8 grams. The vapour density of a gas is

(a) 32

(b) 40

(d) 8

Explanation; (Answer; c)

124.The vapour densities of two gases are in the ratio of 1:3. Their molecular masses are in the ratio of:

(a) 1:3

(b) 1:2

(d) 3:1

Explanation; (Answer; a)

125. An unknown gas has a density 1.96 gL⁻¹. Then gas is

(a) CO

(b) NO₂

(d) SO₂

Explanation; (Answer; c)

Volume of 1 mole of gas at STP = 22.4 L

126. 2.2 g of an unknown gas is present in a container of 1.12 L. Then unknown gas is

(a) CO₂

(b) CO

(d) Both a and c

Explanation; (Answer; d)

The molecular mass of CO₂ (12 + 32) and N₂O (28+16) = 44 amu. So gas unknown gas may be CO₂ or N₂O.

127. 16 g of oxygen has same number of molecules as in

(a) 16 g of CO

(b) 28 g of N₂

(d) 2 g of He

Explanation; (Answer; b)

128. The volume of 3.01 x 10²³ molecules of N₂ gas at STP will be

(a) 3 dm³

(b) 11.2 cm³

(d) 11200 cm³

Explanation; (Answer; d)

129. 8.8 g of CO₂ contains ………… atoms

(a) 0.4 N_A

(b) 0.2 N_A

(d) None of them

Explanation; (Answer; b)

44 g of CO2 contains 3N_A atoms

8.8 g of CO2 contains 3N_A x 8.8/44 = 0.2 N_A atoms

130. The number of atoms in 11.2 L of SO₂ gas at STP is

(a) N_A/2

(b) 3N_A/2

(d) N_A

Explanation; (Answer; b)

No. of atoms = V/V_M x N_A x atomicity = 11.2/22.4 N_A x 3 = 3N_A x ½ = 3N_A /2

131.The total number of electrons present in 18mL of water is

(a) 6.02 x 10²²

(b) 6.02 x 10²³

(d) 6.02 x 10²⁵N_A

Explanation; (Answer; c)

The total number of electrons present in 18 ml of water is 6.023 × 10²⁴.

Number of electron in one molecule of H2O is 2 + 8 = 10.

Density = 1 g/ml

∴18 ml means 18 g

Moles =18 / 18 = 1

Molecules = 6.02 × 10²³

Number of Electrons = 6.02 × 10²³ × 10 = 6.02 × 10²⁴

132. The volume of 1.0 g of hydrogen at NTP is

(a) 2.24 L

(b) 22.4 L

(d) 11.2 L

Explanation; (Answer; d)

1 mole of any gas at NTP occupies 22.4 L.

1 g of hydrogen (H2) constitutes its half mole.

Half mole of any gas at NTP occupies half of 22.4 L i.e. 11.2 L

133. 7.5 grams of a gas occupy 5.6 litres of volume at STP. The gas is

(a) NO

(b) N₂O₄

(d) CO₂

Explanation; (Answer; a)

Molar mass = (V_m / V_g) x mass = 22.4/5.6 x 7.5 = 44 g

Molar mass of NO = 30 gmol^-1

134. How many moles are there in m³ of any gas at NTP?

(a) 44.6

(b) 4.46

(d) 0.0446

Explanation; (Answer; a)

No. of moles of gas at STP = V_g/V_M = 1 m³/0.0224 m³/mol = 44.6 mole

135. no. of moles in 28 kg of silicon.

(a) 0.1

(b) 0.01

(d) 10

Explanation; (Answer; c)

No. of moles = mass/molar = 28 x 10³ g/28 = 103 or 1000 mole.

136. no. of moles in 1000 mg of helium is

(a) 0.25

(b) 0.01

(d) 0.2

Explanation; (Answer; a)

No. of moles = mass/molar = 1000 ÷ 10³ g/4 = ¼ mole or 0.25 mole

137. 49g of phosphoric acid dissolved in excess quantity of water, it will yield ___________ moles of hydrogen ion (H⁺) ___ moles of phosphate (PO₄^3–)

(a) 1.5, 02

(b) 1.5, 01

(d) 1.5, 0.5

Explanation; (Answer; d)

Molar mass of phosphoric acid (H₃PO₄) = 3+31+64 = 98 gmol⁻¹

98g (1 mole) of phosphoric acid (H₃PO₄) ionizes to give 3 mole H⁺ ions and 1 mole phosphate ion (PO₄^3–)

49 (0.5 mol) of phosphoric acid (H₃PO₄) ionizes to give 1.5 mole H⁺ ions and 0.5 mole phosphate ion (PO₄^3–)

138. Number of atoms present in 22.414 liter of hydrogen gas at STP

(a) 3.01x10²³

(b) 2.408x10²³

(d) 1.204x10²⁴

Explanation; (Answer; d)

No of atoms in given volume of gas at STP = atomicity x V_g/V_m)N_A= 2 x (22.414/22.414)N_A= 2 x 1 x N_A = 2N_A

= 2 x 6.02x10²³= 1.204x10²⁴

139. Number of molecules in 18ml of water will be almost

(a) 18x6.02x10²³

(b) 55.6x6.02x10²³

(d) 2x6.02x10²³

Explanation; (Answer; c)

Molecular weight of H₂O = 18 g

Because density of water = 1 g/mL

18 g water = 18 ml water

∵ 18 ml of H₂O contain = 6.02×10²³ molecules of H₂O

140). 0.8 moles of ammonia and 0.8 moles of methane gas have.

(a) Same no. of molecules

(b) Different no of molecules

(d) Same no. of atoms

Explanation; (Answer; a)

Same number of moles contains same number of molecules (and aslo same number of atoms for same atomicity).

141. 14g of silicon and 7g of nitrogen contains:

(a) Nitrogen contains more atoms

(b) Si contains more atoms

(d) Different atoms

Explanation; (Answer; c)

14 g of silicon and 7g of nitrogen both constitutes their half moles. Same number of moles contains same number of particles.

142. One mole of CO₂ contains:

I) 6.02x10²³ atoms of carbon

II) 6.02x10²³ atoms of oxygen

III) 1gram molecule of CO₂

IV) Molecules having 22.4dm³ volume

(a) I and III

(b) II and III

(d) I, III and IV

Explanation; (Answer; d)

One mole of CO₂ contains 1gram molecule of CO₂

One mole of CO₂ contains N_A (6.02x10²³) molecules of CO₂

One mole of CO₂ contains 22.4dm³ volume of CO₂

One mole of CO₂ contains 1 gram atom or 1 mole of C

One mole of CO₂ contains 2 gram molecule or 2 mole of O

One mole of CO₂ contains N_A (6.02x10²³) atoms of C

One mole of CO₂ contains N_A (6.02x10²³) molecules of O₂

143. 180 g of glucose contains atoms of carbon

(a) 0.5N_A

(b) 3N_A

(d) 12N_A

Explanation; (Answer; d)

180 g of glucose constitutes its 1 mole

1 mole glucose = 6 moles of C = 6N_A atoms of C

1 mole glucose = 12 moles of H = 12NA atoms of H

1 mole glucose = 6 moles of C = 6N_A atoms of O

144. No. of ions in one formula unit of sodium chloride

(a) 3 x 6.02 x10²³

(b) 2

(d) 4

Explanation; (Answer; b)

one formula unit of sodium chloride (NaCl) contains 2 mole of ions or 2N_A ions (2 x 6.02 x 10²³ ions).

145.One gram formula of Na₂CO₃ has the number of ions than the number of formula units when dissolved in water.

(a) Twice

(b) Thrice

(d) Same

Explanation; (Answer; b)

One gram formula or 1 mole of gives total three moles of ions (2 Na⁺ and 1 CO₃²⁻) on ionization.

146. The no. of moles of CO₂which contains 6g of carbon

(a) 0.25

(b) 0.50

(d) 1.50

Explanation; (Answer; b)

12 g carbon = 1 mole of CO₂

6 g carbon = 1/12 x 6 = 0.5 mole of CO₂

147. no. of gram ions in 3.9g K⁺

(a) 6.02x10²²

(b) 0.1

(d) 0.01

Explanation; (Answer; a)

No of gram ions or mole = mass/molar mass x N_A = 3.9/39 x N_A = 0.1N_A = 0.1 x 6.02x10²² = 6.02x10²³

148. no. of formula units in 95g MgCl₂.

(a) 6.02x10²³

(b) 12.04x10²³

(d) One

Explanation; (Answer; a)

No of formula units or mole = mass/molar mass x NA = 95/95 x N_A = 1N_A= 6.02x10²³

149.no. of gram atoms of oxygen in 48g ozone

(a) 1

(b) 2

(d) 18.06x10²³

Explanation; (Answer; a)

No of gram atoms or mole = mass/molar mass = 48/48 x N_A = 1

150. 16 gram of methane contains ___________ no. of electrons

(a) 6.02x10²³

(b) 6.02x10²⁴

(d) 6.02x10²⁶

Explanation; (Answer; b)

16 g of methane (CH₄) = 1 mole

1 mole of methane (CH₄) contains total 6+ 4(1) = 10 moles of electrons = 10N_A electrons = 10 x 6.02x10²³ = 6.02x10²⁴

151. 60g CO₃^2– and 24g Mg⁺² have number of ions.

(a) Same

(b) Different

(d) Mg⁺² has more

Explanation; (Answer; a)

No of moles in 60g CO₃^2– = mass/molar mass = 60/60 = 1 mole

No of moles in 24g Mg⁺²= mass/molar mass = 24/24 = 1 mole

Same number of moles contain same number of particles.

152. A molecule of haemoglobin is made up of nearly

(a) 10,1000 atoms

(b) 50,000 atoms

(d) 1500 atoms

Explanation; (Answer; d)

153.Each molecule of haemoglobin is 68000 times heavier than one atom of

(a) C

(b) H

(d) O

Explanation; (Answer; b)

Each molecule of haemoglobin is 68000 times heavier than one atom of H.

154. Relative atomic mass of an element is the mass of an element relative to

(a) 1/12 mass of carbon-12

(b) 1/12 mass of carbon

(d) None of them

Explanation; (Answer; a)

Relative atomic mass of an element is calculated by comparing it with 1/12^th mass of one atom of standard C-12.

45. The relative atomic mass of chlorine is 35.5. What is the mass of 2 moles of chlorine gas?

(a)142 g

(b) 71 g

(d) 18.75 g

Explanation; (Answer; a)

Mass from moles = n x M = 2 x 71 = 142 g

156. How many moles of hydrogen atoms does 3.2 g of methane (CH₄) contain?

(a) 0.02

(b) 0.2

(d) 0.8

Explanation; (Answer; a)

16 g of CH4 contains 4 moles of H

3.2 g of CH4 contains 4/16 x 3.2 = 0.8 mole

157. One mole of C₂H₅OH contains the number of H-atoms

(a) 6.02 x 10²³

(b) 3.61 x 10²⁴

(d) 6.02 x 10²⁴

Explanation; (Answer; b)

There are 6 mole of H atoms in 1 mole of C₂H₅OH

So, there will be 6NA atoms or 3.61 x 10²⁴atoms of H in 1 mole of C₂H₅OH

158. 3.01 x 10²² Ag⁺ ions is present in

(a) 85 g AgNO₃

(b) 0.85 g AgNO₃

(d) 18.5g AgNO₃

Explanation; (Answer; c)

3.01 x 10²² Ag⁺ ions constitutes its 0.05 mole.

6.02 x 10²³ Ag⁺ ions = 170 g (1 mole) of AgNO₃

3.01 x 10²² Ag⁺ ions = 3.01 x 10²² /6.02 x 10²³ x 170 g (1 mole) of AgNO₃= 8.5 g AgNO₃

159. The mass of one mole of proton is

(a) 1.008 g

(b) 0.184 g

(d) 1.008 mg

Explanation; (Answer; a)

Mass of 1 proton = 1.672×10⁻²⁴g

Mass of one mole (6.02 x 10²³) protons = 1.672×10⁻²⁴× 6.022×10²³= 1.0068784 g ≈ 1.007 g

160. 0.5 mole of CH₄ and 0.5 mole of SO₂gases have equal

(a) Total number of atoms

(b) Number of molecules

(d) volume

Explanation; (Answer; b)

For same number of moles of different substances, there are same number of molecules.

161. A beaker containing 180 g of water contains:

(a) 6.02 x 10²³molecules

(b) 6.02 x 10²⁴ molecules

(d) 12.01 x 10²³ molecules

Explanation; (Answer; b)

180 g of water constitutes its 10 moles.

No of molecules in 10 moles = n N_A = 10N_A = 10 x 6.02 x 10²³ = 60.2 x 10²³ = 6.02 x 10²⁴ molecules

162. The present scale of atomic mass is based on 1 amu being equal to the mass of:

(a) 1 hydrogen atom

(b) 1/10 of O-16

(d) 1/16 of O-16

Explanation; (Answer; c)

An atomic mass unit (abbreviated as a.m.u) is a physiochemical constant and it is defined as one twelfth (1/12) of the mass of a single atom (the most abundant lightest isotope) of carbon-12 (¹²C). [Because the mass of an atom’s electrons is negligible compared with the mass of its proton and neutrons, defining 1 amu as 1/12 the mass of a ¹²C atom means that both protons and neutrons have a mass of almost exactly 1 amu]. 1 amu is equivalent to 1.66 x 10^–24 g OR 1.66 x 10^–27 kg (which is approximately equal to the mass of one hydrogen atom). Prior to 1960, the amu was defined in terms of the mass of a ¹⁶O-isotope (1.6599 x 10^–27 kg).

163. Chlorine atom and chloride ions

(a) Have same number of protons

(b) Are allotropes of chlorine

(d) Are chemically identical

Explanation; (Answer; a)

Atom and its ion have same number of protons but different number of electrons. Atom and its ion are chemically different.

164.This one of the following pairs has the same number of molecules

(a) 10 g H₂ and 10 g CH₄

(b) 10 g H₂ and 50 g CH₄

(d) 10 g H₂ and 80 g CH₄

Explanation; (Answer; c)

Same number of moles contains same number of molecules.

(a) 10 g H₂ and 10 g CH₄ constitutes 5 mole H₂ and 0.625 mole CH₄

(b) 10 g H₂ and 50 g CH₄ constitutes 5 mole H₂ and 3.125 mole CH₄

(d) 10 g H₂ and 80 g CH₄ constitutes 5 mole H₂ and 5 mole CH₄

MCQs Set # 2 (Chapter # 1 …. Introduction to Chemistry)

1. To determine the molecular formula of glucose, its empirical formula is multiplied by:

2. The prefix 10¹⁸

3. What is the ratio of the weight of water formed to the weight of oxygen used in the formation of water?

4. One mole of carbon-12 has a mass of:

5. What is the mass of one mole of iodine molecule?

6. Which of the following is classified as a monoatomic element?

7. The present scale of atomic mass is based on 1 amu being equal to the mass of:

8. The mass of a single hydrogen atom is:

9. 11.2 dm³ of nitrogen gas at STP weighs:

10.The mass of one atom of coal is:

Explanatory Answers of MCQs Set # 2(Chapter # 1 …… Introduction to Chemistry)

1. Solution

For glucose, value of integer n is 6.

2. Solution

Prefixes (Multiples & Fractions) used with S.I. and Metric Units

Prefixes	Multiples	Prefixes	Fractions
Dec(D)	10¹(Ten)	deci (d)	10^–1
Hecto(H)	10²(Hundred)	centi (c)	10^–2
kilo (k)	10³(Thousand)	milli (m)	10^–3
Mega(M)	10⁶(Million)	micro (µ)	10^–6
Giga (G)	10⁹(Billion)	nano (n)	10^–9
Tera (T)	10¹²(Trillion)	pico (p)	10^–12
Peta (P)	10¹⁵(Quadrillion)	femto(f/fm)	10^–15
Exa (E)	10¹⁸(Quintillion)	atto (a)	10^–18
Zetta (Z)	10²¹(Sextillion)	zepto (z)	10^–21
Yottta (Y)	10²⁴(Septillion)	yacto (y)	10^–24

3. Solution

4. Solution

Mass of one mole of carbon-12 is 12 g or 12/1000 = 0.012 kg

5. Solution

6. Solution

Noble gases (e.g. He, Ne, Ar, Kr, Xe and Rn) are classified as monoatomic elements.

7. Solution

An atomic mass unit (abbreviated as a.m.u) is a physiochemical constant and it is defined as one twelfth (1/12) of the mass of a single atom (the most abundant lightest isotope) of carbon-12 (¹²C).

[Because the mass of an atom’s electrons is negligible compared with the mass of its proton and neutrons, defining 1 amu as 1/12 the mass of a ¹²C atom means that both protons and neutrons have a mass of almost exactly 1 amu].

1 amu is equivalent to 1.66 x 10^–24 g OR 1.66 x 10^–27 kg (which is approximately equal to the mass of one hydrogen atom). Prior to 1960, the amu was defined in terms of the mass of a ¹⁶O-isotope (1.6599 x 10^–27 kg).

8. Solution

9. Solution

22.4 dm³ of nitrogen gas at STP weighs 28 g

10. Solution

11. Solution

The weight of one mole of KAl(SO₄)₂.12H₂O is equal to its gram formula mass as it is an ionic compound.

KAl(SO₄)₂.12H₂O = 39 + 27 + 2(32) + 8(16) + 12(18) = 474 g

12. Solution

13. Solution

14. Solution

32 g of oxygen is present in 1 mole of CO₂

1 g of oxygen is present in 1/32 mole of CO₂

8 g oxygen is present in 1/32 x 8 mole of CO₂ = 0.25 mole of CO₂

15. Solution

Mass of one electron is 9.11 x 10^-31 kg

Mass of one mole (6.02 x 10²³) electron is = 9.11 x 0^-31 x 6.02 x 10²³ = 54.84 x 10^‒8 kg or 5.5 x 10^‒7 kg

Mass of one mole of electron in mg = 5.5 x 10^-7 kg x 1 x 10⁶ = 5.5 x 10^‒1 mg or 0.55 mg

16. Solution

8 of O constitutes its half mole.

17. Solution

18. Solution

Protons in 1 mole of CaCO₃ = Z of Ca + Z of C + Z of O x 3 Þ 20 + 6 + 24 = 50 moles of protons

100 g of CaCO₃ contains 50 moles of protons

1 g of CaCO₃ contains 50/100 moles of protons

10 of CaCO₃ contains = 50/100 x 10 x 6.02 x 10²³ moles of protons = 3.0115 x 10²⁴ protons

19. Solution

2 g of oxygen constitutes its 0.125 mole

0.125 mole of oxygen contains number of atoms equal to 0.125 x 6.02 x 10²³ = 7.525 x 10²²atoms

0.5 g of hydrogen constitutes its 0.25 mole

2.3 g of sodium constitutes its 0.1 mole

7 g of nitrogen constitutes its 0.5 mole

4 g of sulphur constitutes its 0.125 mole

4 g of sulphur and 2 g of oxygen constitutes same number of moles (i.e. 0.125 mole), therefore, they contain same number of atoms.

20. Solution

MCQs on Mole Concept Set # 3

Chemistry by Inam Jazbi

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XI MCQs on Mole Concept with Explanatory Answers

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Diagonal Relationship of Representative Elements

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