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Multiple Choice Questions, Stoichiometry

Choose the correct answers.

i. The branch of chemistry which deals with the calculations based on balance chemical equations is called …………

a) Environmental Chemistry

b) Physical Chemistry

c) Stoichiometry

d) Industrial Chemistry

ii) The mass of an atom compared with the mass of one atom of C12 is called …………….. of that atom.

a)      one mole

b)      gram atomic mass

c)       atomic number

d)      relative atomic mass

iii) Which of the following is not true for a mole?

a) It is a counting unit

b) It is the gram atomic or gram formula mass of a substance.

c) It contains 6.023× 1023 particles.

d) It contains different number of particles for different substances.

Explanation: It has been established by the work of Italian scientist, Amedeo Avogadro, that one mole of any substance consists of same number of particles, 6.023 × 1023, (option c). By particle we mean atoms or molecules.

Note 1: Mole is a counting unit. While dealing with balanced chemical reactions, we count the number of moles of one substance reacting with the number of moles of another substance giving a particular number of moles of the product (again a substance). So, mole is a unit of amount, like kilogram. So option (a) is true.

Note 2: By definition, one mole is the atomic or formula mass expressed in grams. So option (b) is true.

iv. What is the mass (in g) of 5 moles of water (H2O).

a) 90 g

b) 36 g

c) 18 g

d) 100 g

Explanation: Do you know what the formula mass of H2O is? Calculate it. It is 18 g. Therefore, if we take 18 g of H2O, it will make one mole of water. Therefore, 5 mole of H2O will be 18 × 5 = 90 g.

v. The number of molecules 22 g of CO2 is ……..

a) 6.023×1023

b) 3.011×1023

c) 6.021×1021

d) 6.023×1022

Solution: Calculate the formula mass of CO2. It is 12 + 2×16 = 12+32 = 44 g.

Now, 44 g of CO2 consists of 6.023×1023 molecules of CO2. Therefore, 22 (= 44/2) g of CO2 will contain ½×(6.023×1023) = 3.011 × 1023 number of molecules.

vi. Which of the following conditions of temperature and pressure are standard conditions (STP)?

(i) 00C, 1 atm

(ii) 273 K, 1 atm

(iii) 273 K, 760 mm Hg

 

a) (i) only

b) (ii) only

c) (i) and (ii) only

d) All (i), (ii), (iii)

Explanation: All the three options are equal.

vii. The molar volume of SO2 gas at STP is ……..

a) 64 dm3

b) 24 dm3

c) 22.4 dm3

d) 100 cm3

Explanation: Molar volume of all gases is 22.4 dm3.

viii. The percentage of Ca in CaCO3 is ……

a) 12 %

b) 100 %

c) 48 %

d) 40 %

Solution: Formula mass of CaCO3 is = 40 + 12 +  3(16) = 40 + 12 + 48 = 100

Now, 100 g of CaCO3 consists of 40 g of Ca. Therefore, the percentage of Ca in CaCO3 is 40.

ix. Given the equation CO2 (g) + C(s) → 2CO(g). Which of the following equivalences is not correct in the reaction.

  1. 1 mol CO2 ≅ 2 mol of CO
  2. 1 mol C ≅ 56 g CO
  3. 44 g CO2 ≅ 28 g CO
  4. 44 g CO2 ≅ 12 g C

Explanation: The given reaction says, 1 mol of CO2 reacts with 1 mol of C to yield 2 mol of CO. Now consider,

  1. a) 1 mol of CO2 (on LHS) is congruent to 2 mol of CO (left). Hence correct.
  2. b) 1 mol of CO = 12 + 16 = 28 g, therefore, 2 mol of CO is 56 g. Hence correct.
  3. c) In the reaction, 1 mol of CO2 is equivalent to 2 mol of CO. But 1 mol of CO2 = 12 + 32 = 44 g. And, 2 mol of CO = 2 (12 + 16) = 2 × 28 = 56 g. Therefore, 44 g CO2 ≅ 56 g CO. Thus the option is not correct.
  4. d) Do this correct option yourself.

(x) Theoretical yield is always less than actual yield, because

  1. Some product is lost in the experiment
  2. Reversible reaction may occur
  3. Errors are made in weighing in the reactants or the products.
  4. The given statement is not correct.

Explanation: Actually, theoretical yield is always more than the actual yield.

(xi) Actual yield will reach the ideal (theoretical) value if % yield of the reaction is ………

a) 50 %

b) 90 %

c) 100 %

d) 1 %

Explanation: When all reactants undergo the reaction and consumed, the product will be maximum (ideal).

(xii) The largest number of molecules is present in ………

a) 44 g of CO2

b) 98 g of H2SO4

c) 36 g of H2O

d) 180 g of C6H12O6

Explanation: Using the Avogadro’s number, we calculate the number of molecules of the given quantity of the substances.

  1. a) 1 mol of CO2 = 12 + 32 = 44 g = 6.023 × 1023 (Given value)
  2. b) 1 mol of H2SO2 = 2 + 32 + 64 = 98 g = 6.023 × 1023 (Given value)
  3. c) 1 mole of H2O = 2 + 16 = 18 g = 6.023 × 1023 molecules ==> 36 g of H2O = 2 × 6.023 × 1023 = 12.046 × 1023
  4. d) 1 mol of C6H12O6 = 72 + 12 + 96 = 180 g = 6.023 × 1023 molecules (given value).

Compare all values, we see 36 g of H20 has maximum number of molecules (double).

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