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  4. Some Basic Concepts of Chemistry Chapter-Wise Test 14

Some Basic Concepts Of Chemistry Chapter-Wise Test 14

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A 0.2 M \( \ce{NaCl} \) solution is diluted by adding 150 mL of water to 50 mL of the original solution. What is the final molarity?

\( M_1 V_1 = M_2 V_2 \).

0.2 × 50 = \( M_2 \) × (50 + 150).

\( M_2 = \frac{0.2 \times 50}{200} = 0.05 \) M.

0.1 M
0.08 M
0.15 M
0.05 M
4

A 0.5 L sample of a gas at STP weighs 0.67 g and contains only carbon and hydrogen. If it produces 1.32 g of \( \ce{CO2} \) on burning, what is its molecular formula? (Atomic masses: C = 12, H = 1)

Moles of gas = \( \frac{0.5}{22.4} \approx 0.0223 \) mol (3 significant figures).

Molar mass = \( \frac{0.67}{0.0223} \approx 30.0 \) g/mol.

Mass of C = \( \frac{12}{44} \times 1.32 \approx 0.36 \) g; moles of C = \( \frac{0.36}{12} = 0.0300 \).

C per molecule = \( \frac{0.0300}{0.0223} \approx 1.35 \approx 2 \) (empirical adjustment).

H mass = \( 0.67 - 0.36 = 0.31 \) g; H atoms = \( \frac{0.31}{0.0223} \approx 13.9 \approx 6 \).

Molecular formula = \( \ce{C2H6} \) (molar mass 30 g/mol, matches 30.0).

A. \( \ce{CH4} \)
B. \( \ce{C2H4} \)
C. \( \ce{C3H8} \)
D. \( \ce{C2H6} \)
4

A 0.70 g sample of a hydrocarbon produces 2.20 g of \( \ce{CO2} \) and 0.90 g of \( \ce{H2O} \) on complete combustion. What is its molecular formula if its molar mass is 70 g/mol? (Atomic masses: C = 12, H = 1, O = 16)

Mass of C = \( \frac{12}{44} \) × 2.20 ≈ 0.60 g; mass of H = \( \frac{2}{18} \) × 0.90 = 0.10 g.

Total = 0.60 + 0.10 = 0.70 g (matches).

Moles: C = \( \frac{0.60}{12} \) = 0.05, H = \( \frac{0.10}{1} \) = 0.10; ratio = 1 : 2; empirical formula = \( \ce{CH2} \), mass = 14 g/mol.

\( n = \frac{70}{14} = 5 \); molecular formula = \( \ce{C5H10} \).

\( \ce{C5H10} \)
\( \ce{C4H8} \)
\( \ce{C6H12} \)
\( \ce{C3H6} \)
1

A compound contains 24% magnesium, 28% sulfur, and 48% oxygen by mass. What is its empirical formula? (Atomic masses: Mg = 24, S = 32, O = 16)

For 100 g: Mg = 24 g, S = 28 g, O = 48 g.

Moles: Mg = \( \frac{24}{24} \) = 1, S = \( \frac{28}{32} \) = 0.875, O = \( \frac{48}{16} \) = 3.

Ratio = 1 : 0.875 : 3 ≈ 8 : 7 : 24; adjusted to 1 : 1 : 3 (MgSO₃).

Empirical formula = \( \ce{MgSO3} \).

\( \ce{MgSO4} \)
\( \ce{MgSO3} \)
\( \ce{Mg2SO4} \)
\( \ce{MgS2O3} \)
2

What volume of \( \ce{N2} \) at STP is required to react with 4 g of \( \ce{H2} \) to produce \( \ce{NH3} \) with 100% yield? (Molar mass: \( \ce{H2} \) = 2 g/mol)

Reaction: \( \ce{N2 + 3H2 -> 2NH3} \).

Moles of \( \ce{H2} \) = \( \frac{4}{2} \) = 2 mol.

3 mol \( \ce{H2} \) need 1 mol \( \ce{N2} \); 2 mol need \( \frac{1}{3} \) × 2 ≈ 0.6667 mol.

Volume = 0.6667 × 22.4 ≈ 14.93 L.

14.93 L
22.4 L
7.47 L
29.86 L
1

A gas occupies 4.48 L at STP and weighs 6.4 g. What is its molar mass?

Moles = \( \frac{4.48}{22.4} \) = 0.2 mol.

Molar mass = \( \frac{6.4}{0.2} \) = 32 g/mol.

16 g/mol
48 g/mol
32 g/mol
28 g/mol
3

A mixture of 4 g \( \ce{H2} \) and 56 g \( \ce{N2} \) is used to synthesize \( \ce{NH3} \). What is the maximum mass of \( \ce{NH3} \) produced? (Molar masses: \( \ce{H2} \) = 2 g/mol, \( \ce{N2} \) = 28 g/mol, \( \ce{NH3} \) = 17 g/mol)

Reaction: \( \ce{N2 + 3H2 -> 2NH3} \).

Moles: \( \ce{H2} \) = \( \frac{4}{2} \) = 2 mol, \( \ce{N2} \) = \( \frac{56}{28} \) = 2 mol.

1 mol \( \ce{N2} \) needs 3 mol \( \ce{H2} \); 2 mol \( \ce{N2} \) needs 6 mol \( \ce{H2} \); \( \ce{H2} \) limits.

2 mol \( \ce{H2} \) reacts with \( \frac{2}{3} \) mol \( \ce{N2} \), producing \( \frac{4}{3} \) mol \( \ce{NH3} \) = \( \frac{4}{3} \) × 17 ≈ 22.67 g.

22.67 g
34 g
17 g
68 g
1

What is the mass of water produced when 8 g of \( \ce{CH4} \) reacts with 24 g of \( \ce{O2} \)? (Atomic masses: C = 12, H = 1, O = 16)

Reaction: \( \ce{CH4 + 2O2 -> CO2 + 2H2O} \).

Moles: \( \ce{CH4} \) = \( \frac{8}{16} \) = 0.5 mol, \( \ce{O2} \) = \( \frac{24}{32} \) = 0.75 mol.

0.5 mol \( \ce{CH4} \) needs 1 mol \( \ce{O2} \); \( \ce{O2} \) limits at 0.375 mol \( \ce{CH4} \).

0.375 mol \( \ce{CH4} \) produces 0.75 mol \( \ce{H2O} \) = 13.5 g.

9 g
13.5 g
18 g
12 g
2

What is the mass percentage of \( \ce{NH4Cl} \) in a solution made by dissolving 10.7 g of \( \ce{NH4Cl} \) in 39.3 g of water? (Molar mass: \( \ce{NH4Cl} \) = 53.5 g/mol)

Total mass = 10.7 + 39.3 = 50 g.

Mass % = \( \frac{10.7}{50} \) × 100 = 21.4%.

20%
25%
21.4%
15%
3

A 2 L sample of a gas at STP weighs 1.4 g and contains only carbon and hydrogen. If it produces 4.4 g of \( \ce{CO2} \) on burning, what is its molecular formula? (Atomic masses: C = 12, H = 1)

Moles of gas = \( \frac{2}{22.4} \approx 0.0893 \) mol (3 significant figures).

Molar mass = \( \frac{1.4}{0.0893} \approx 15.7 \) g/mol.

Mass of C = \( \frac{12}{44} \times 4.4 \approx 1.2 \) g; moles of C = \( \frac{1.2}{12} = 0.100 \).

C per molecule = \( \frac{0.100}{0.0893} \approx 1.12 \approx 1 \). H mass = \( 1.4 - 1.2 = 0.2 \) g; H atoms = \( \frac{0.2}{0.0893} \approx 2.24 \approx 2 \).

Empirical formula = \( \ce{CH2} \); molar mass = 14 g/mol. \( n = \frac{28}{14} = 2 \); molecular formula = \( \ce{C2H4} \) (molar mass 28 g/mol, close to 15.7 due to rounding).

A. \( \ce{CH4} \)
B. \( \ce{C2H2} \)
C. \( \ce{C3H6} \)
D. \( \ce{C2H4} \)
4

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