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  4. Some Basic Concepts of Chemistry Chapter-Wise Test 10

Some Basic Concepts Of Chemistry Chapter-Wise Test 10

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A solution contains 20 ppm of \( \ce{K2SO4} \) by mass in water. What is its molarity if the density is 1 g/mL? (Molar mass: \( \ce{K2SO4} \) = 174 g/mol)

20 ppm = 20 g \( \ce{K2SO4} \) in 10⁶ g solution.

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

Moles = \( \frac{20}{174} \) ≈ 0.1149 mol; molarity = \( \frac{0.1149}{1000} \) ≈ 1.149 × 10⁻⁴ M.

2.0 × 10⁻⁴ M
1.0 × 10⁻⁴ M
1.149 × 10⁻⁴ M
5.0 × 10⁻⁵ M
3

What is the mole fraction of \( \ce{CH3OH} \) in a solution containing 16 g of \( \ce{CH3OH} \) and 72 g of \( \ce{H2O} \)? (Molar masses: \( \ce{CH3OH} \) = 32 g/mol, \( \ce{H2O} \) = 18 g/mol)

Moles of \( \ce{CH3OH} \) = \( \frac{16}{32} \) = 0.5 mol; moles of \( \ce{H2O} \) = \( \frac{72}{18} \) = 4 mol.

Total moles = 0.5 + 4 = 4.5.

Mole fraction = \( \frac{0.5}{4.5} \) ≈ 0.111.

0.2
0.111
0.25
0.15
2

A hydrocarbon (C and H only) weighing 0.26 g produces 0.88 g of \( \ce{CO2} \) and 0.18 g of \( \ce{H2O} \) on complete combustion. What is its molecular formula if its molar mass is 26 g/mol? (Atomic masses: C = 12, H = 1, O = 16)

Mass of C = \( \frac{12}{44} \) × 0.88 ≈ 0.24 g; mass of H = \( \frac{2}{18} \) × 0.18 = 0.02 g.

Total mass = 0.24 + 0.02 = 0.26 g (matches).

Moles: C = \( \frac{0.24}{12} \) = 0.02, H = \( \frac{0.02}{1} \) = 0.02; ratio = 1 : 1; empirical formula = \( \ce{CH} \), mass = 13 g/mol.

\( n = \frac{26}{13} = 2 \); molecular formula = \( \ce{C2H2} \).

\( \ce{C2H2} \)
\( \ce{CH4} \)
\( \ce{C2H4} \)
\( \ce{C3H6} \)
1

How many grams of \( \ce{Fe2O3} \) are required to produce 16.8 g of \( \ce{Fe} \) with excess \( \ce{CO} \)? (Molar masses: \( \ce{Fe2O3} \) = 160 g/mol, Fe = 56 g/mol)

Reaction: \( \ce{Fe2O3 + 3CO -> 2Fe + 3CO2} \).

Moles of Fe = \( \frac{16.8}{56} \) = 0.3 mol.

2 mol Fe from 1 mol \( \ce{Fe2O3} \); 0.3 mol from 0.15 mol.

Mass = 0.15 × 160 = 24 g.

12 g
24 g
48 g
32 g
2

A 1.5 L sample of a gas at STP contains 0.75 g of carbon and 0.25 g of hydrogen. What is its molecular formula? (Atomic masses: C = 12, H = 1)

Moles of gas = \( \frac{1.5}{22.4} \) ≈ 0.067 mol; total mass = 0.75 + 0.25 = 1 g.

Molar mass = \( \frac{1}{0.067} \) ≈ 14.93 g/mol.

Moles: C = \( \frac{0.75}{12} \) ≈ 0.0625, H = \( \frac{0.25}{1} \) = 0.25; per molecule: C = \( \frac{0.0625}{0.067} \) ≈ 1, H = \( \frac{0.25}{0.067} \) ≈ 4.

Molecular formula = \( \ce{CH4} \) (molar mass ≈ 16, close to 14.93).

\( \ce{C2H2} \)
\( \ce{C2H4} \)
\( \ce{CH2} \)
\( \ce{CH4} \)
4

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 = \( \ce{C3H6} \).

\( \ce{C3H6} \)
\( \ce{C2H4} \)
\( \ce{C4H8} \)
\( \ce{CH2} \)
1

A solution contains 10 ppm of \( \ce{K2SO4} \) by mass in water. What is its molality? (Molar mass: \( \ce{K2SO4} \) = 174 g/mol)

10 ppm = 10 g \( \ce{K2SO4} \) in 10⁶ g water.

Moles = \( \frac{10}{174} \) ≈ 0.0575 mol.

Mass of solvent = 10⁶ g = 1000 kg.

Molality = \( \frac{0.0575}{1000} \) ≈ 5.75 × 10⁻⁵ m.

5.0 × 10⁻⁵ m
6.0 × 10⁻⁵ m
5.75 × 10⁻⁵ m
4.5 × 10⁻⁵ m
3

What volume of \( \ce{O2} \) at STP is required to burn 5 g of \( \ce{C2H5OH} \) completely to \( \ce{CO2} \) and \( \ce{H2O} \)? (Molar mass: \( \ce{C2H5OH} \) = 46 g/mol)

Reaction: \( \ce{C2H5OH + 3O2 -> 2CO2 + 3H2O} \).

Moles of \( \ce{C2H5OH} \) = \( \frac{5}{46} \) ≈ 0.1087 mol.

1 mol \( \ce{C2H5OH} \) needs 3 mol \( \ce{O2} \); 0.1087 mol needs 0.3261 mol.

Volume = 0.3261 × 22.4 ≈ 7.3 L.

7.3 L
2.43 L
14.6 L
22.4 L
1

How many grams of \( \ce{HCl} \) are needed to react with 10 g of \( \ce{CaCO3} \) to produce 4.4 g of \( \ce{CO2} \)? (Molar masses: \( \ce{CaCO3} \) = 100 g/mol, \( \ce{HCl} \) = 36.5 g/mol, \( \ce{CO2} \) = 44 g/mol)

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

Moles of \( \ce{CO2} \) = \( \frac{4.4}{44} \) = 0.1 mol; needs 0.1 mol \( \ce{CaCO3} \) (10 g).

0.1 mol \( \ce{CaCO3} \) requires 0.2 mol \( \ce{HCl} \) = 0.2 × 36.5 = 7.3 g.

3.65 g
7.3 g
14.6 g
10 g
2

A 500 mL solution of \( \ce{KNO3} \) has a molarity of 0.4 M and a density of 1.01 g/mL. What is the mass percentage of \( \ce{KNO3} \)? (Molar mass: \( \ce{KNO3} \) = 101 g/mol)

Moles = 0.4 × 0.5 = 0.2 mol; mass of \( \ce{KNO3} \) = 0.2 × 101 = 20.2 g.

Mass of solution = 500 × 1.01 = 505 g.

Mass % = \( \frac{20.2}{505} \) × 100 ≈ 4%.

2%
4%
6%
8%
2

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