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

Some Basic Concepts Of Chemistry Chapter-Wise Test 4

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A 0.3 M \( \ce{HCl} \) solution is diluted by adding 400 mL of water to 100 mL of the original solution. What is the final molarity?

\( M_1 V_1 = M_2 V_2 \).

0.3 × 100 = \( M_2 \) × (100 + 400).

\( M_2 = \frac{0.3 \times 100}{500} = 0.06 \) M.

0.15 M
0.09 M
0.12 M
0.06 M
4

What is the mass percentage of \( \ce{KNO3} \) in a solution made by dissolving 10.1 g of \( \ce{KNO3} \) in 39.9 g of water? (Molar mass: \( \ce{KNO3} \) = 101 g/mol)

Total mass = 10.1 + 39.9 = 50 g.

Mass % = \( \frac{10.1}{50} \) × 100 = 20.2%.

18%
22%
20.2%
25%
3

What is the mass of \( \ce{PbSO4} \) produced when 33.1 g of \( \ce{Pb(NO3)2} \) reacts with excess \( \ce{H2SO4} \)? (Molar masses: \( \ce{Pb(NO3)2} \) = 331 g/mol, \( \ce{PbSO4} \) = 303 g/mol)

Reaction: \( \ce{Pb(NO3)2 + H2SO4 -> PbSO4 + 2HNO3} \).

Moles of \( \ce{Pb(NO3)2} \) = \( \frac{33.1}{331} \) = 0.1 mol.

1 mol \( \ce{Pb(NO3)2} \) produces 1 mol \( \ce{PbSO4} \); mass = 0.1 × 303 = 30.3 g.

15.15 g
45.45 g
22.7 g
30.3 g
4

A solution of \( \ce{HNO3} \) has a molarity of 1.5 M and a density of 1.05 g/mL. What is its molality? (Molar mass: \( \ce{HNO3} \) = 63 g/mol)

Mass of 1 L solution = 1000 × 1.05 = 1050 g.

Mass of \( \ce{HNO3} \) = 1.5 × 63 = 94.5 g.

Mass of water = 1050 - 94.5 = 955.5 g = 0.9555 kg.

Molality = \( \frac{1.5}{0.9555} \) ≈ 1.57 m.

1.57 m
1.43 m
1.75 m
2.0 m
1

How many significant figures are present in the result of \( \frac{0.0894 \times 298.15}{0.0821} \)?

0.0894 (3 sig figs), 298.15 (5 sig figs), 0.0821 (3 sig figs).

Result ≈ 324.67; limited to 3 sig figs (325).

2
4
3
5
3

A 0.1 M \( \ce{KOH} \) solution is diluted by adding 200 mL of water to 100 mL of the original solution. What is the final molarity?

\( M_1 V_1 = M_2 V_2 \).

0.1 × 100 = \( M_2 \) × (100 + 200).

\( M_2 = \frac{0.1 \times 100}{300} \) ≈ 0.0333 M.

0.05 M
0.1 M
0.02 M
0.0333 M
4

What is the mass of \( \ce{CaCO3} \) required to react with 50 mL of 2 M \( \ce{HCl} \)? (Molar mass: \( \ce{CaCO3} \) = 100 g/mol)

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

Moles of \( \ce{HCl} \) = 2 × 0.05 = 0.1 mol.

0.1 mol \( \ce{HCl} \) needs 0.05 mol \( \ce{CaCO3} \) = 0.05 × 100 = 5 g.

10 g
2.5 g
7.5 g
5 g
4

What is the mass of \( \ce{KClO3} \) required to produce 2.24 L of \( \ce{O2} \) at STP with 50% decomposition efficiency? (Molar mass: \( \ce{KClO3} \) = 122.5 g/mol)

Reaction: \( \ce{2KClO3 -> 2KCl + 3O2} \).

Moles of \( \ce{O2} \) = \( \frac{2.24}{22.4} \) = 0.1 mol.

3 mol \( \ce{O2} \) from 2 mol \( \ce{KClO3} \); 0.1 mol needs \( \frac{2}{3} \) × 0.1 ≈ 0.0667 mol.

With 50% efficiency, \( \ce{KClO3} \) = \( \frac{0.0667}{0.5} \) ≈ 0.1333 mol; mass = 0.1333 × 122.5 ≈ 16.33 g.

8.17 g
12.25 g
24.5 g
16.33 g
4

A compound contains 32.43% sodium, 22.52% sulfur, and 45.05% oxygen by mass. What is its empirical formula? (Atomic masses: Na = 23, S = 32, O = 16)

For 100 g: Na = 32.43 g, S = 22.52 g, O = 45.05 g.

Moles: Na = \( \frac{32.43}{23} \) ≈ 1.41, S = \( \frac{22.52}{32} \) ≈ 0.704, O = \( \frac{45.05}{16} \) ≈ 2.82.

Ratio = 2 : 1 : 4; empirical formula = \( \ce{Na2SO4} \).

\( \ce{NaSO2} \)
\( \ce{Na2SO4} \)
\( \ce{Na2S2O3} \)
\( \ce{NaSO3} \)
2

What volume of \( \ce{O2} \) at STP is required to burn 11 g of \( \ce{C2H6} \) completely to \( \ce{CO2} \) and \( \ce{H2O} \)? (Molar mass: \( \ce{C2H6} \) = 30 g/mol)

Reaction: \( \ce{2C2H6 + 7O2 -> 4CO2 + 6H2O} \).

Moles of \( \ce{C2H6} \) = \( \frac{11}{30} \) ≈ 0.3667 mol.

2 mol \( \ce{C2H6} \) need 7 mol \( \ce{O2} \); 0.3667 mol need \( \frac{7}{2} \) × 0.3667 ≈ 1.2833 mol.

Volume = 1.2833 × 22.4 ≈ 28.75 L.

28.75 L
14.38 L
22.4 L
33.6 L
1

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