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  4. Some Basic Concepts of Chemistry Chapter-Wise Test 5

Some Basic Concepts Of Chemistry Chapter-Wise Test 5

Correct answer Carries: 4.

Wrong Answer Carries: -1.

How many grams of \( \ce{Na2SO4} \) are produced when 8 g of \( \ce{NaOH} \) reacts with excess \( \ce{H2SO4} \)? (Molar masses: \( \ce{NaOH} \) = 40 g/mol, \( \ce{Na2SO4} \) = 142 g/mol)

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

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

2 mol \( \ce{NaOH} \) produce 1 mol \( \ce{Na2SO4} \); 0.2 mol produce 0.1 mol.

Mass = 0.1 × 142 = 14.2 g.

7.1 g
14.2 g
28.4 g
10.6 g
1

A gas occupies 3.36 L at STP and weighs 4.8 g. What is its molar mass?

Moles = \( \frac{3.36}{22.4} \) = 0.15 mol.

Molar mass = \( \frac{4.8}{0.15} \) = 32 g/mol.

16 g/mol
48 g/mol
32 g/mol
24 g/mol
3

A mixture of 10 g \( \ce{C} \) and 24 g \( \ce{O2} \) is ignited to form \( \ce{CO2} \). What is the mass of \( \ce{CO2} \) produced? (Atomic masses: C = 12, O = 16)

Reaction: \( \ce{C + O2 -> CO2} \).

Moles: C = \( \frac{10}{12} \) ≈ 0.8333 mol, \( \ce{O2} \) = \( \frac{24}{32} \) = 0.75 mol.

1 mol C needs 1 mol \( \ce{O2} \); 0.75 mol \( \ce{O2} \) limits, reacts with 0.75 mol C.

0.75 mol \( \ce{CO2} \) formed = 0.75 × 44 = 33 g.

33 g
44 g
22 g
11 g
1

What volume of \( \ce{N2} \) at STP is required to react with 6 g of \( \ce{H2} \) to produce \( \ce{NH3} \) with 60% yield? (Molar mass: \( \ce{H2} \) = 2 g/mol)

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

Moles of \( \ce{H2} \) = \( \frac{6}{2} \) = 3 mol.

3 mol \( \ce{H2} \) need 1 mol \( \ce{N2} \); actual \( \ce{N2} \) = \( \frac{1}{0.6} \) ≈ 1.67 mol.

Volume = 1.67 × 22.4 ≈ 37.41 L.

37.41 L
22.4 L
11.2 L
44.8 L
1

What volume of \( \ce{CO2} \) at STP is produced when 10 g of \( \ce{C4H10} \) is burned completely? (Molar mass: \( \ce{C4H10} \) = 58 g/mol)

Reaction: \( \ce{2C4H10 + 13O2 -> 8CO2 + 10H2O} \).

Moles of \( \ce{C4H10} \) = \( \frac{10}{58} \) ≈ 0.1724 mol.

2 mol \( \ce{C4H10} \) produce 8 mol \( \ce{CO2} \); 0.1724 mol produce \( \frac{8}{2} \) × 0.1724 ≈ 0.6896 mol.

Volume = 0.6896 × 22.4 ≈ 15.45 L.

7.73 L
23.19 L
11.58 L
15.45 L
4

A 400 mL solution contains 14.6 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 \( \ce{HCl} \) = \( \frac{14.6}{36.5} \) = 0.4 mol.

Volume = 400 mL = 0.4 L.

Molarity = \( \frac{0.4}{0.4} \) = 1 M.

0.5 M
1 M
1.5 M
2 M
2

A 0.42 g sample of a hydrocarbon produces 1.32 g of \( \ce{CO2} \) and 0.54 g of \( \ce{H2O} \) on complete combustion. What is its molecular formula if its molar mass is 42 g/mol? (Atomic masses: C = 12, H = 1, O = 16)

Mass of C = \( \frac{12}{44} \) × 1.32 ≈ 0.36 g; mass of H = \( \frac{2}{18} \) × 0.54 = 0.06 g.

Total = 0.36 + 0.06 = 0.42 g (matches).

Moles: C = \( \frac{0.36}{12} \) = 0.03, H = \( \frac{0.06}{1} \) = 0.06; ratio = 1 : 2; empirical formula = \( \ce{CH2} \), mass = 14 g/mol.

\( n = \frac{42}{14} = 3 \); molecular formula = \( \ce{C3H6} \).

\( \ce{C3H6} \)
\( \ce{C2H4} \)
\( \ce{C4H8} \)
\( \ce{C5H10} \)
1

A 0.84 g sample of a hydrocarbon produces 2.64 g of \( \ce{CO2} \) and 1.08 g of \( \ce{H2O} \) on complete combustion. What is its molecular formula if its molar mass is 84 g/mol? (Atomic masses: C = 12, H = 1, O = 16)

Mass of C = \( \frac{12}{44} \) × 2.64 ≈ 0.72 g; mass of H = \( \frac{2}{18} \) × 1.08 = 0.12 g.

Total = 0.72 + 0.12 = 0.84 g (matches).

Moles: C = \( \frac{0.72}{12} \) = 0.06, H = \( \frac{0.12}{1} \) = 0.12; ratio = 1 : 2; empirical formula = \( \ce{CH2} \), mass = 14 g/mol.

\( n = \frac{84}{14} = 6 \); molecular formula = \( \ce{C6H12} \).

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

What mass of \( \ce{N2} \) reacts with 6 g of \( \ce{H2} \) to form \( \ce{NH3} \), if \( \ce{H2} \) is the limiting reagent? (Molar masses: \( \ce{N2} \) = 28 g/mol, \( \ce{H2} \) = 2 g/mol)

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

Moles of \( \ce{H2} \) = \( \frac{6}{2} \) = 3 mol.

1 mol \( \ce{N2} \) needs 3 mol \( \ce{H2} \); 3 mol \( \ce{H2} \) needs 1 mol \( \ce{N2} \) = 28 g.

14 g
56 g
28 g
42 g
3

A solution contains 10 ppm of \( \ce{SO2} \) by mass in water. What is the molarity if the density of the solution is 1 g/mL? (Molar mass: \( \ce{SO2} \) = 64 g/mol)

10 ppm = 10 g \( \ce{SO2} \) in 10⁶ g solution.

Volume = \( \frac{10⁶}{1} \) = 10⁶ mL = 1000 L.

Moles = \( \frac{10}{64} \) ≈ 0.15625 mol; molarity = \( \frac{0.15625}{1000} \) ≈ 1.5625 × 10⁻⁴ M.

1.0 × 10⁻⁴ M
2.0 × 10⁻⁴ M
1.5625 × 10⁻⁴ M
5.0 × 10⁻⁴ M
3

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