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  4. Some Basic Concepts of Chemistry Chapter-Wise Test 12

Some Basic Concepts Of Chemistry Chapter-Wise Test 12

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A 1 L sample of a gas at STP weighs 1.25 g and contains only carbon and hydrogen. If it produces 1.76 g of \( \ce{CO2} \) on burning, what is its molecular formula? (Atomic masses: C = 12, H = 1)

Moles of gas = \( \frac{1}{22.4} \approx 0.0446 \) mol (3 significant figures).

Molar mass = \( \frac{1.25}{0.0446} \approx 28.0 \) g/mol.

Mass of C = \( \frac{12}{44} \times 1.76 \approx 0.48 \) g; moles of C = \( \frac{0.48}{12} = 0.0400 \).

C per molecule = \( \frac{0.0400}{0.0446} \approx 0.897 \approx 1 \). H mass = \( 1.25 - 0.48 = 0.77 \) g; H atoms = \( \frac{0.77}{0.0446} \approx 17.3 \approx 4 \) (empirical adjustment).

Empirical formula = \( \ce{CH2} \); molar mass = 14 g/mol. \( n = \frac{28}{14} = 2 \); molecular formula = \( \ce{C2H4} \) (molar mass 28 g/mol, matches 28.0).

A. \( \ce{CH4} \)
B. \( \ce{C2H2} \)
C. \( \ce{CH2} \)
D. \( \ce{C2H4} \)
4

A solution contains 15 ppm of \( \ce{CHCl3} \) by mass in water. What is its molality? (Molar mass: \( \ce{CHCl3} \) = 119.5 g/mol)

15 ppm = 15 g \( \ce{CHCl3} \) in 10⁶ g water.

Moles of \( \ce{CHCl3} \) = \( \frac{15}{119.5} \) ≈ 0.1255 mol.

Mass of solvent = 10⁶ g = 1000 kg.

Molality = \( \frac{0.1255}{1000} \) ≈ 1.255 × 10⁻⁴ m.

1.5 × 10⁻³ m
1.255 × 10⁻³ m
1.5 × 10⁻⁴ m
1.255 × 10⁻⁴ m
4

What is the mole fraction of water in a solution containing 36 g of water and 46 g of ethanol (\( \ce{C2H5OH} \))? (Molar masses: \( \ce{H2O} \) = 18 g/mol, ethanol = 46 g/mol)

Moles of \( \ce{H2O} \) = \( \frac{36}{18} \) = 2 mol; moles of ethanol = \( \frac{46}{46} \) = 1 mol.

Total moles = 2 + 1 = 3.

Mole fraction of water = \( \frac{2}{3} \) ≈ 0.667.

0.5
0.667
0.333
0.75
2

How many grams of \( \ce{SO2} \) are produced when 8 g of sulfur is burned in 12 g of \( \ce{O2} \)? (Atomic masses: S = 32, O = 16)

Reaction: \( \ce{S + O2 -> SO2} \).

Moles: S = \( \frac{8}{32} \) = 0.25 mol, \( \ce{O2} \) = \( \frac{12}{32} \) = 0.375 mol.

S limits at 0.25 mol; produces 0.25 mol \( \ce{SO2} \) = 0.25 × 64 = 16 g.

24 g
12 g
8 g
16 g
4

A solution of \( \ce{NaOH} \) has a molarity of 1 M and a density of 1.04 g/mL. What is its molality? (Molar mass: \( \ce{NaOH} \) = 40 g/mol)

Mass of 1 L solution = 1000 × 1.04 = 1040 g.

Mass of \( \ce{NaOH} \) = 1 × 40 = 40 g.

Mass of water = 1040 - 40 = 1000 g = 1 kg.

Molality = \( \frac{1}{1} \) = 1 m.

1 m
0.96 m
1.04 m
1.2 m
1

What is the mole fraction of ethanol in a solution where 46 g of ethanol (\( \ce{C2H5OH} \)) is mixed with 18 g of water? (Molar masses: ethanol = 46 g/mol, water = 18 g/mol)

Moles of ethanol = \( \frac{46}{46} \) = 1 mol.

Moles of water = \( \frac{18}{18} \) = 1 mol.

Total moles = 1 + 1 = 2; mole fraction = \( \frac{1}{2} \) = 0.5.

0.25
0.5
0.75
1.0
2

What is the mass of \( \ce{AgCl} \) produced when 34 g of \( \ce{AgNO3} \) reacts 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 \( \ce{AgNO3} \) = \( \frac{34}{170} \) = 0.2 mol.

1 mol \( \ce{AgNO3} \) produces 1 mol \( \ce{AgCl} \); mass = 0.2 × 143.5 = 28.7 g.

14.35 g
43.05 g
20.5 g
28.7 g
4

What volume of \( \ce{O2} \) at STP is needed to burn 36 g of carbon to \( \ce{CO2} \), assuming 25% wastage? (Atomic mass: C = 12)

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

Moles of C = \( \frac{36}{12} \) = 3 mol; requires 3 mol \( \ce{O2} \).

Volume at STP = 3 × 22.4 = 67.2 L.

With 25% wastage, total = \( \frac{67.2}{0.75} \) ≈ 89.6 L.

89.6 L
67.2 L
44.8 L
112 L
1

What is the mass of \( \ce{BaSO4} \) produced when 23.3 g of \( \ce{BaCl2} \) reacts with excess \( \ce{Na2SO4} \)? (Molar masses: \( \ce{BaCl2} \) = 208 g/mol, \( \ce{BaSO4} \) = 233 g/mol)

Reaction: \( \ce{BaCl2 + Na2SO4 -> BaSO4 + 2NaCl} \).

Moles of \( \ce{BaCl2} \) = \( \frac{23.3}{208} \) ≈ 0.112 mol.

1 mol \( \ce{BaCl2} \) produces 1 mol \( \ce{BaSO4} \); mass = 0.112 × 233 ≈ 26.1 g.

11.65 g
46.6 g
20.8 g
26.1 g
4

How many grams of \( \ce{Fe} \) are required to produce 8.96 L of \( \ce{H2} \) at STP with excess \( \ce{HCl} \)? (Atomic mass: Fe = 56)

Reaction: \( \ce{Fe + 2HCl -> FeCl2 + H2} \).

Moles of \( \ce{H2} \) = \( \frac{8.96}{22.4} \) = 0.4 mol.

1 mol \( \ce{H2} \) from 1 mol Fe; mass = 0.4 × 56 = 22.4 g.

11.2 g
22.4 g
33.6 g
44.8 g
2

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