1. Home
    2. >
  2. Neet Chemistry Class 11 Mock Test Papers
    3. >
  3. Chemistry Chapter Wise Test Papers
    4. >
  4. Some Basic Concepts of Chemistry Chapter-Wise Test 16

Some Basic Concepts Of Chemistry Chapter-Wise Test 16

Correct answer Carries: 4.

Wrong Answer Carries: -1.

How many grams of \( \ce{ZnCl2} \) are required to produce 6.5 g of \( \ce{Zn} \) with excess \( \ce{HCl} \)? (Molar masses: \( \ce{ZnCl2} \) = 136 g/mol, Zn = 65 g/mol)

Reaction: \( \ce{ZnCl2 + H2 -> Zn + 2HCl} \) (reverse context with excess HCl).

Moles of Zn = \( \frac{6.5}{65} \) = 0.1 mol.

1 mol Zn from 1 mol \( \ce{ZnCl2} \); mass = 0.1 × 136 = 13.6 g.

6.8 g
13.6 g
27.2 g
20.4 g
2

A compound has a molar mass of 180 g/mol and contains 40% carbon, 6.67% hydrogen, and 53.33% oxygen. What is its molecular formula? (Atomic masses: C = 12, H = 1, O = 16)

For 100 g: C = 40 g, H = 6.67 g, O = 53.33 g.

Moles: C = \( \frac{40}{12} \) ≈ 3.33, H = \( \frac{6.67}{1} \) ≈ 6.67, O = \( \frac{53.33}{16} \) ≈ 3.33.

Ratio = 1 : 2 : 1; empirical formula = \( \ce{CH2O} \), mass = 30 g/mol.

\( n = \frac{180}{30} = 6 \); molecular formula = \( \ce{C6H12O6} \).

\( \ce{C3H6O3} \)
\( \ce{C4H8O4} \)
\( \ce{C6H12O6} \)
\( \ce{CH2O} \)
3

A gas occupies 0.56 L at STP and weighs 0.64 g. What is its molar mass?

Moles = \( \frac{0.56}{22.4} \) = 0.025 mol.

Molar mass = \( \frac{0.64}{0.025} \) = 25.6 g/mol.

32 g/mol
16 g/mol
25.6 g/mol
28 g/mol
3

How many grams of \( \ce{AlCl3} \) are required to produce 5.4 g of \( \ce{Al} \) with excess \( \ce{HCl} \)? (Molar masses: \( \ce{AlCl3} \) = 133.5 g/mol, Al = 27 g/mol)

Reaction: \( \ce{2AlCl3 + 3H2 -> 2Al + 6HCl} \) (contextual reverse).

Moles of Al = \( \frac{5.4}{27} \) = 0.2 mol.

2 mol Al from 2 mol \( \ce{AlCl3} \); 0.2 mol from 0.2 mol.

Mass = 0.2 × 133.5 = 26.7 g.

13.35 g
40.05 g
20.03 g
26.7 g
4

How many grams of \( \ce{NaOH} \) are required to produce 11.7 g of \( \ce{NaCl} \) with excess \( \ce{HCl} \)? (Molar masses: \( \ce{NaOH} \) = 40 g/mol, \( \ce{NaCl} \) = 58.5 g/mol)

Reaction: \( \ce{NaOH + HCl -> NaCl + H2O} \).

Moles of \( \ce{NaCl} \) = \( \frac{11.7}{58.5} \) = 0.2 mol.

1 mol \( \ce{NaCl} \) from 1 mol \( \ce{NaOH} \); mass = 0.2 × 40 = 8 g.

4 g
12 g
8 g
16 g
3

How many grams of \( \ce{NaCl} \) are produced when 10.6 g of \( \ce{Na2CO3} \) reacts with excess \( \ce{HCl} \)? (Molar masses: \( \ce{Na2CO3} \) = 106 g/mol, \( \ce{NaCl} \) = 58.5 g/mol)

Reaction: \( \ce{Na2CO3 + 2HCl -> 2NaCl + CO2 + H2O} \).

Moles of \( \ce{Na2CO3} \) = \( \frac{10.6}{106} \) = 0.1 mol.

1 mol \( \ce{Na2CO3} \) produces 2 mol \( \ce{NaCl} \); 0.1 mol produces 0.2 mol.

Mass = 0.2 × 58.5 = 11.7 g.

5.85 g
11.7 g
23.4 g
17.55 g
1

A 400 mL solution of \( \ce{HCl} \) has a molarity of 0.25 M and a density of 1.01 g/mL. What is the mass percentage of \( \ce{HCl} \)? (Molar mass: \( \ce{HCl} \) = 36.5 g/mol)

Moles = 0.25 × 0.4 = 0.1 mol; mass of \( \ce{HCl} \) = 0.1 × 36.5 = 3.65 g.

Mass of solution = 400 × 1.01 = 404 g.

Mass % = \( \frac{3.65}{404} \) × 100 ≈ 0.904%.

1.5%
0.904%
0.5%
2.0%
2

What is the mole fraction of \( \ce{CH3OH} \) in a solution containing 8 g of \( \ce{CH3OH} \) and 54 g of \( \ce{H2O} \)? (Molar masses: \( \ce{CH3OH} \) = 32 g/mol, \( \ce{H2O} \) = 18 g/mol)

Moles of \( \ce{CH3OH} \) = \( \frac{8}{32} \) = 0.25 mol; moles of \( \ce{H2O} \) = \( \frac{54}{18} \) = 3 mol.

Total moles = 0.25 + 3 = 3.25.

Mole fraction = \( \frac{0.25}{3.25} \) ≈ 0.0769.

0.1
0.0769
0.25
0.05
2

A 300 mL solution contains 11.7 g of \( \ce{NaCl} \) and has a density of 1.01 g/mL. What is its molarity? (Molar mass: \( \ce{NaCl} \) = 58.5 g/mol)

Moles of \( \ce{NaCl} \) = \( \frac{11.7}{58.5} \) = 0.2 mol.

Volume = 300 mL = 0.3 L.

Molarity = \( \frac{0.2}{0.3} \) ≈ 0.667 M.

0.5 M
0.667 M
1.0 M
0.33 M
2

A 2 L sample of a gas at STP weighs 1.4 g and contains only carbon and hydrogen. If it produces 4.4 g of \( \ce{CO2} \) on burning, what is its molecular formula? (Atomic masses: C = 12, H = 1)

Moles of gas = \( \frac{2}{22.4} \approx 0.0893 \) mol (3 significant figures).

Molar mass = \( \frac{1.4}{0.0893} \approx 15.68 \) g/mol, rounded to 15.7 g/mol (3 significant figures).

Mass of C = \( \frac{12}{44} \times 4.4 \approx 1.2 \) g; moles of C = \( \frac{1.2}{12} = 0.10 \) mol.

C atoms per molecule = \( \frac{0.10}{0.0893} \approx 1.12 \approx 1 \). H mass = \( 1.4 - 1.2 = 0.2 \) g; H atoms = \( \frac{0.2}{0.0893} \approx 2.24 \approx 4 \) (empirical adjustment).

Empirical formula = \( \ce{CH4} \); molar mass = 16 g/mol, close to 15.7 g/mol, confirming molecular formula \( \ce{CH4} \).

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

Performance Summary

Score:

Category Details
Total Attempts:
Total Skipped:
Total Wrong Answers:
Total Correct Answers:
Time Taken:
Average Time Taken per Question:
Accuracy:
0