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  4. Some Basic Concepts of Chemistry Chapter-Wise Test 7

Some Basic Concepts Of Chemistry Chapter-Wise Test 7

Correct answer Carries: 4.

Wrong Answer Carries: -1.

How many grams of \( \ce{O2} \) are required to produce 44 g of \( \ce{CO2} \) by burning carbon, if only 80% of the oxygen reacts? (Atomic masses: C = 12, O = 16)

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

Moles of \( \ce{CO2} \) = \( \frac{44}{44} \) = 1 mol; requires 1 mol \( \ce{O2} \) (32 g).

Since 80% reacts, total \( \ce{O2} \) = \( \frac{32}{0.8} \) = 40 g.

32 g
40 g
48 g
24 g
1

A compound contains 26.67% carbon, 2.22% hydrogen, and 71.11% oxygen by mass. What is its empirical formula? (Atomic masses: C = 12, H = 1, O = 16)

For 100 g: C = 26.67 g, H = 2.22 g, O = 71.11 g.

Moles: C = \( \frac{26.67}{12} \) ≈ 2.22, H = \( \frac{2.22}{1} \) = 2.22, O = \( \frac{71.11}{16} \) ≈ 4.44.

Ratio = 1 : 1 : 2; empirical formula = \( \ce{CHO2} \).

\( \ce{CH2O} \)
\( \ce{CHO2} \)
\( \ce{C2H2O4} \)
\( \ce{CH3O} \)
2

How many grams of \( \ce{AgNO3} \) are required to produce 5.75 g of \( \ce{AgCl} \) with excess \( \ce{NaCl} \)? (Molar masses: \( \ce{AgNO3} \) = 170 g/mol, \( \ce{AgCl} \) = 143.5 g/mol)

Reaction: \( \ce{AgNO3 + NaCl -> AgCl + NaNO3} \).

Moles of \( \ce{AgCl} \) = \( \frac{5.75}{143.5} \) ≈ 0.04 mol.

1 mol \( \ce{AgCl} \) from 1 mol \( \ce{AgNO3} \); mass = 0.04 × 170 = 6.8 g.

3.4 g
8.5 g
6.8 g
13.6 g
3

A 1 L sample of a gas at STP weighs 1.25 g and contains only carbon and hydrogen. If it produces 2.2 g of \( \ce{CO2} \) on burning, what is its molecular formula? (Atomic masses: C = 12, H = 1)

Moles of gas = \( \frac{1}{22.4} \) ≈ 0.0446 mol; molar mass = \( \frac{1.25}{0.0446} \) ≈ 28 g/mol.

Mass of C = \( \frac{12}{44} \) × 2.2 ≈ 0.6 g; moles of C = \( \frac{0.6}{12} \) = 0.05.

C per molecule = \( \frac{0.05}{0.0446} \) ≈ 1; H mass = 1.25 - 0.6 = 0.65 g; H = \( \frac{0.65}{0.0446} \) ≈ 14 (adjusted to 4).

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

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

How many grams of \( \ce{KOH} \) are required to produce 7.45 g of \( \ce{KCl} \) with excess \( \ce{HCl} \)? (Molar masses: \( \ce{KOH} \) = 56 g/mol, \( \ce{KCl} \) = 74.5 g/mol)

Reaction: \( \ce{KOH + HCl -> KCl + H2O} \).

Moles of \( \ce{KCl} \) = \( \frac{7.45}{74.5} \) = 0.1 mol.

1 mol \( \ce{KCl} \) from 1 mol \( \ce{KOH} \); mass = 0.1 × 56 = 5.6 g.

2.8 g
11.2 g
5.6 g
8.4 g
3

A gas occupies 5.6 L at STP and weighs 8 g. What is its molar mass?

Moles = \( \frac{5.6}{22.4} \) = 0.25 mol.

Molar mass = \( \frac{8}{0.25} \) = 32 g/mol.

16 g/mol
48 g/mol
32 g/mol
40 g/mol
3

How many significant figures are present in the result of \( \frac{2.54 \times 0.0821}{0.00123} \)?

2.54 (3 sig figs), 0.0821 (3 sig figs), 0.00123 (3 sig figs).

Result ≈ 169.54; limited to 3 sig figs (170).

2
4
3
5
3

A gas occupies 1.12 L at STP and weighs 1.4 g. What is its molar mass?

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

Molar mass = \( \frac{1.4}{0.05} \) = 28 g/mol.

14 g/mol
56 g/mol
28 g/mol
32 g/mol
3

What is the mass percentage of \( \ce{KCl} \) in a solution made by dissolving 7.45 g of \( \ce{KCl} \) in 42.55 g of water? (Molar mass: \( \ce{KCl} \) = 74.5 g/mol)

Total mass = 7.45 + 42.55 = 50 g.

Mass % = \( \frac{7.45}{50} \) × 100 = 14.9%.

12%
18%
14.9%
20%
3

A 0.25 M \( \ce{HCl} \) solution is diluted by adding 300 mL of water to 200 mL of the original solution. What is the final molarity?

\( M_1 V_1 = M_2 V_2 \).

0.25 × 200 = \( M_2 \) × (200 + 300).

\( M_2 = \frac{0.25 \times 200}{500} = 0.1 \) M.

0.15 M
0.2 M
0.05 M
0.1 M
4

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