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 mol CO2 ≅ 2 mol of CO
- 1 mol C ≅ 56 g CO
- 44 g CO2 ≅ 28 g CO
- 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,
- a) 1 mol of CO2 (on LHS) is congruent to 2 mol of CO (left). Hence correct.
- b) 1 mol of CO = 12 + 16 = 28 g, therefore, 2 mol of CO is 56 g. Hence correct.
- 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.
- d) Do this correct option yourself.
(x) Theoretical yield is always less than actual yield, because
- Some product is lost in the experiment
- Reversible reaction may occur
- Errors are made in weighing in the reactants or the products.
- 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.
- a) 1 mol of CO2 = 12 + 32 = 44 g = 6.023 × 1023 (Given value)
- b) 1 mol of H2SO2 = 2 + 32 + 64 = 98 g = 6.023 × 1023 (Given value)
- c) 1 mole of H2O = 2 + 16 = 18 g = 6.023 × 1023 molecules ==> 36 g of H2O = 2 × 6.023 × 1023 = 12.046 × 1023
- 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).
Fantastic mcqs
Marwa
Thank you so much!
My chemistry is weak please help me
– Study and discuss with your friend.
– Don’t start next topic until you have fully understood the current one.
– Start studying from your 9-grade chemistry book.
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