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  4. Some Basic Concepts of Chemistry Chapter-Wise Test 6

Some Basic Concepts Of Chemistry Chapter-Wise Test 6

Correct answer Carries: 4.

Wrong Answer Carries: -1.

What volume of \( \ce{CO2} \) at STP is produced when 7 g of \( \ce{C2H2} \) is burned completely? (Molar mass: \( \ce{C2H2} \) = 26 g/mol)

Reaction: \( \ce{2C2H2 + 5O2 -> 4CO2 + 2H2O} \).

Moles of \( \ce{C2H2} \) = \( \frac{7}{26} \) ≈ 0.2692 mol.

2 mol \( \ce{C2H2} \) produce 4 mol \( \ce{CO2} \); 0.2692 mol produce 0.5384 mol.

Volume = 0.5384 × 22.4 ≈ 12.06 L.

6.03 L
18.09 L
9.02 L
12.06 L
4

What volume of \( \ce{CO2} \) at STP is produced when 11.2 g of \( \ce{C5H12} \) is burned completely? (Molar mass: \( \ce{C5H12} \) = 72 g/mol)

Reaction: \( \ce{C5H12 + 8O2 -> 5CO2 + 6H2O} \).

Moles of \( \ce{C5H12} \) = \( \frac{11.2}{72} \) ≈ 0.1556 mol.

1 mol \( \ce{C5H12} \) produces 5 mol \( \ce{CO2} \); 0.1556 mol produces 0.7778 mol.

Volume = 0.7778 × 22.4 ≈ 17.42 L.

8.71 L
26.13 L
13.08 L
17.42 L
4

How many grams of \( \ce{CuSO4} \) are required to produce 3.2 g of \( \ce{Cu} \) with excess \( \ce{Zn} \)? (Molar masses: \( \ce{CuSO4} \) = 159.5 g/mol, Cu = 64 g/mol)

Reaction: \( \ce{CuSO4 + Zn -> Cu + ZnSO4} \).

Moles of Cu = \( \frac{3.2}{64} \) = 0.05 mol.

1 mol \( \ce{Cu} \) from 1 mol \( \ce{CuSO4} \); mass = 0.05 × 159.5 = 7.975 g ≈ 8 g.

4 g
15.95 g
8 g
12 g
3

A 200 mL solution contains 7.3 g of \( \ce{HCl} \) and has a density of 1.02 g/mL. What is its molarity? (Molar mass: \( \ce{HCl} \) = 36.5 g/mol)

Moles of \( \ce{HCl} \) = \( \frac{7.3}{36.5} \) = 0.2 mol.

Volume = 200 mL = 0.2 L.

Molarity = \( \frac{0.2}{0.2} \) = 1 M.

0.5 M
1 M
2 M
1.5 M
2

A gas occupies 1.12 L at STP and weighs 1.6 g. What is its molar mass?

Moles = \( \frac{1.12}{22.4} \) = 0.05 mol.

Molar mass = \( \frac{1.6}{0.05} \) = 32 g/mol.

16 g/mol
64 g/mol
32 g/mol
28 g/mol
3

What is the mole fraction of \( \ce{H2O} \) in a solution containing 36 g of \( \ce{H2O} \) and 46 g of \( \ce{C2H5OH} \)? (Molar masses: \( \ce{H2O} \) = 18 g/mol, \( \ce{C2H5OH} \) = 46 g/mol)

Moles of \( \ce{H2O} \) = \( \frac{36}{18} \) = 2 mol; moles of \( \ce{C2H5OH} \) = \( \frac{46}{46} \) = 1 mol.

Total moles = 2 + 1 = 3.

Mole fraction = \( \frac{2}{3} \) ≈ 0.667.

0.5
0.667
0.333
0.75
2

How many grams of \( \ce{H2SO4} \) are required to neutralize 8 g of \( \ce{NaOH} \) completely? (Molar masses: \( \ce{H2SO4} \) = 98 g/mol, \( \ce{NaOH} \) = 40 g/mol)

Reaction: \( \ce{H2SO4 + 2NaOH -> Na2SO4 + 2H2O} \).

Moles of \( \ce{NaOH} \) = \( \frac{8}{40} \) = 0.2 mol.

2 mol \( \ce{NaOH} \) need 1 mol \( \ce{H2SO4} \); 0.2 mol needs 0.1 mol.

Mass = 0.1 × 98 = 9.8 g.

4.9 g
9.8 g
19.6 g
14.7 g
1

A 3 L sample of a gas at STP weighs 3.75 g and contains only carbon and hydrogen. If it produces 5.28 g of \( \ce{CO2} \) on burning, what is its molecular formula? (Atomic masses: C = 12, H = 1)

Moles of gas = \( \frac{3}{22.4} \) ≈ 0.1339 mol; molar mass = \( \frac{3.75}{0.1339} \) ≈ 28 g/mol.

Mass of C = \( \frac{12}{44} \) × 5.28 ≈ 1.44 g; moles of C = \( \frac{1.44}{12} \) = 0.12.

C per molecule = \( \frac{0.12}{0.1339} \) ≈ 1; H mass = 3.75 - 1.44 = 2.31 g; H = \( \frac{2.31}{0.1339} \) ≈ 17 (adjusted to 4).

Molecular formula = \( \ce{C2H4} \) (molar mass 28).

\( \ce{CH4} \)
\( \ce{C2H2} \)
\( \ce{C3H6} \)
\( \ce{C2H4} \)
4

What is the mass of \( \ce{SO3} \) produced when 16 g of \( \ce{SO2} \) reacts with 8 g of \( \ce{O2} \)? (Molar masses: \( \ce{SO2} \) = 64 g/mol, \( \ce{SO3} \) = 80 g/mol, \( \ce{O2} \) = 32 g/mol)

Reaction: \( \ce{2SO2 + O2 -> 2SO3} \).

Moles: \( \ce{SO2} \) = \( \frac{16}{64} \) = 0.25 mol, \( \ce{O2} \) = \( \frac{8}{32} \) = 0.25 mol.

2 mol \( \ce{SO2} \) need 1 mol \( \ce{O2} \); 0.25 mol \( \ce{SO2} \) need 0.125 mol \( \ce{O2} \) (available).

0.25 mol \( \ce{SO2} \) produces 0.25 mol \( \ce{SO3} \) = 0.25 × 80 = 20 g.

16 g
24 g
10 g
20 g
4

How many liters of \( \ce{CO2} \) at STP are produced when 10 g of \( \ce{CaCO3} \) reacts with excess \( \ce{HCl} \) with 80% efficiency? (Molar mass: \( \ce{CaCO3} \) = 100 g/mol)

Reaction: \( \ce{CaCO3 + 2HCl -> CaCl2 + CO2 + H2O} \).

Moles of \( \ce{CaCO3} \) = \( \frac{10}{100} \) = 0.1 mol; theoretical \( \ce{CO2} \) = 0.1 mol.

Actual \( \ce{CO2} \) = 0.1 × 0.8 = 0.08 mol; volume = 0.08 × 22.4 = 1.792 L.

2.24 L
1.792 L
1.12 L
0.896 L
2

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