🧪 Atoms and Molecules – MCQs for Class 9 (NCERT)

Section 4: Numerical Problems & Applications (Q76–Q100)

Q76. How many moles are present in 12 g of oxygen gas (O₂)?

A. 0.25 mol
B. 0.375 mol
C. 0.5 mol
D. 0.75 mol

✅ Correct Answer: B. 0.375 mol

Explanation:

• Molar mass of O₂ = 32 g.
• Moles = 12/32 = 0.375 mol.
• A. 0.25 mol → Wrong.
• B. Correct.
• C. 0.5 mol → Wrong.
• D. 0.75 mol → Wrong.

Q77. Number of molecules in 4 g hydrogen gas (H₂):

A. 6.022×10236.022 \times 10^{23}
B. 1.204×10241.204 \times 10^{24}
C. 2.011×10232.011 \times 10^{23}
D. 3.011×10233.011 \times 10^{23}

✅ Correct Answer: A. 6.022×10236.022 \times 10^{23}

Explanation:

• Molar mass H₂ = 2 g.
• 4 g = 2 moles → molecules = 2 × 6.022×10236.022 \times 10^{23}.
  👉 Correction: Answer = B. 1.204×10241.204 \times 10^{24}.

Q78. What is the mass of 0.5 mole of CH₄?

A. 8 g
B. 16 g
C. 32 g
D. 64 g

✅ Correct Answer: B. 16 g

Explanation:

• Molar mass CH₄ = 16 g.
• 0.5 mol = 0.5 × 16 = 8 g.
  👉 Correction: Answer is A. 8 g.

Q79. How many atoms are present in 22 g of CO₂?

A. 6.022×10236.022 \times 10^{23}
B. 9.033×10239.033 \times 10^{23}
C. 3.011×10233.011 \times 10^{23}
D. 1.204×10241.204 \times 10^{24}

✅ Correct Answer: B. 9.033×10239.033 \times 10^{23}

Explanation:

• Molar mass CO₂ = 44 g.
• 22 g = 0.5 mol.
• Molecules = 0.5 × 6.022×10236.022 \times 10^{23} = 3.011×10233.011 \times 10^{23}.
• Each molecule = 3 atoms → total = 9.033×10239.033 \times 10^{23}.

Q80. How many grams of NaOH are present in 0.25 mole?

A. 5 g
B. 10 g
C. 20 g
D. 40 g

✅ Correct Answer: C. 10 g

Explanation:

• Molar mass NaOH = 40 g.
• Mass = 0.25 × 40 = 10 g.

Q81. Which of the following contains 6.022 × 10²³ atoms?

A. 1 mole of O₂
B. 1 mole of H₂
C. 1 mole of He
D. 1 mole of H₂O

✅ Correct Answer: C. 1 mole of He

Explanation:

• A. O₂ → 1 mole O₂ = 2 × Avogadro atoms.
• B. H₂ → 1 mole H₂ = 2 × Avogadro atoms.
• C. He → Correct, monoatomic gas.
• D. H₂O → 3 × Avogadro atoms.

Q82. Calculate the number of molecules in 11 g of CO₂.

A. 1.505×10231.505 \times 10^{23}
B. 3.011×10233.011 \times 10^{23}
C. 6.022×10236.022 \times 10^{23}
D. 4.5×10234.5 \times 10^{23}

✅ Correct Answer: A. 1.505×10231.505 \times 10^{23}

Explanation:

• Moles = 11/44 = 0.25.
• Molecules = 0.25 × 6.022×10236.022 \times 10^{23} = 1.505×10231.505 \times 10^{23}.

Q83. Mass of 2.5 × 10²³ molecules of O₂ is:

A. 4 g
B. 8 g
C. 16 g
D. 32 g

✅ Correct Answer: B. 8 g

Explanation:

• Molecules in 1 mol O₂ = 6.022×10236.022 \times 10^{23}.
• Given = 2.5×10232.5 \times 10^{23}.
• Moles = (2.5/6.022) ≈ 0.415 mol.
• Mass = 0.415 × 32 ≈ 13.3 g.
  👉 Correction: none of the given options match correctly; expected ~13 g. Likely exam typo.

Q84. Number of atoms in 0.5 mole of oxygen gas (O₂):

A. 3.011×10233.011 \times 10^{23}
B. 6.022×10236.022 \times 10^{23}
C. 1.204×10241.204 \times 10^{24}
D. 2.408×10242.408 \times 10^{24}

✅ Correct Answer: C. 6.022×10236.022 \times 10^{23}

Explanation:

• 0.5 mol O₂ = 0.5 × Avogadro molecules.
• Each O₂ has 2 atoms → total = 6.022×10236.022 \times 10^{23}.

Q85. Volume of 0.5 mole of CO₂ at STP is:

A. 11.2 L
B. 22.4 L
C. 44.8 L
D. 5.6 L

✅ Correct Answer: A. 11.2 L

Explanation:

• 1 mol gas = 22.4 L.
• 0.5 mol = 11.2 L.

Q86. How many protons are present in 1 atom of Mg (Z = 12)?

A. 6
B. 12
C. 24
D. 18

✅ Correct Answer: B. 12

Explanation:

• Atomic number = number of protons.
• Mg (Z=12) → 12 protons.

Q87. What is the mass of 5 moles of H₂O?

A. 18 g
B. 45 g
C. 90 g
D. 120 g

✅ Correct Answer: C. 90 g

Explanation:

• 1 mol H₂O = 18 g.
• 5 mol = 5 × 18 = 90 g.

Q88. Which of the following has maximum number of molecules?

A. 9 g H₂O
B. 5.6 L O₂ at STP
C. 2 g H₂
D. 44 g CO₂

✅ Correct Answer: B. 5.6 L O₂ at STP

Explanation:

• A. 9 g H₂O = 0.5 mol → 0.5 × Avogadro molecules.
• B. 5.6 L O₂ = 0.25 mol → 0.25 × Avogadro molecules.
• C. 2 g H₂ = 1 mol → Avogadro molecules → maximum.
  👉 Correction: C. 2 g H₂ is correct.

Q89. Calculate number of moles in 27 g of Al (M = 27).

A. 1 mol
B. 0.5 mol
C. 2 mol
D. 3 mol

✅ Correct Answer: A. 1 mol

Explanation:

• Mass / molar mass = 27/27 = 1 mol.

Q90. Calculate mass of 0.25 mole of nitrogen gas (N₂).

A. 7 g
B. 14 g
C. 28 g
D. 56 g

✅ Correct Answer: B. 7 g

Explanation:

• Molar mass N₂ = 28 g.
• 0.25 mol × 28 = 7 g.

Q91. The total number of atoms in 4.4 g of CO₂ is:

A. 6.022×10226.022 \times 10^{22}
B. 1.806×10231.806 \times 10^{23}
C. 2.408×10232.408 \times 10^{23}
D. 9.033×10239.033 \times 10^{23}

✅ Correct Answer: B. 1.806×10231.806 \times 10^{23}

Explanation:

• 4.4 g = 0.1 mol CO₂.
• Molecules = 0.1 × Avogadro = 6.022×10226.022 \times 10^{22}.
• Atoms per molecule = 3.
• Total = 3 × 6.022×10226.022 \times 10^{22} = 1.806×10231.806 \times 10^{23}.

Q92. How many grams of O₂ are required to completely burn 2 g H₂?

A. 8 g
B. 16 g
C. 32 g
D. 4 g

✅ Correct Answer: A. 16 g

Explanation:

• Reaction: 2H₂ + O₂ → 2H₂O.
• 4 g H₂ needs 32 g O₂.
• 2 g H₂ needs 16 g O₂.

Q93. How many moles of oxygen are there in 88 g CO₂?

A. 1 mol
B. 2 mol
C. 4 mol
D. 6 mol

✅ Correct Answer: C. 4 mol

Explanation:

• 88 g CO₂ = 2 mol.
• Each CO₂ has 2 O atoms → 4 mol oxygen atoms.

Q94. The number of particles in 0.25 mole of any substance is:

A. 1.505×10231.505 \times 10^{23}
B. 3.011×10233.011 \times 10^{23}
C. 6.022×10236.022 \times 10^{23}
D. 7.5×10227.5 \times 10^{22}

✅ Correct Answer: A. 1.505×10231.505 \times 10^{23}

Explanation:

• 0.25 × Avogadro = 1.505×10231.505 \times 10^{23}.

Q95. The number of atoms in 1 g H atom is:

A. 6.022×10236.022 \times 10^{23}
B. 3.011×10233.011 \times 10^{23}
C. 1.204×10241.204 \times 10^{24}
D. 2.011×10232.011 \times 10^{23}

✅ Correct Answer: B. 6.022×10236.022 \times 10^{23}

Explanation:

• 1 g H = 1 mol H atoms.
• Atoms = Avogadro’s number.

Q96. Calculate the number of molecules in 28 g of N₂.

A. 6.022×10236.022 \times 10^{23}
B. 3.011×10233.011 \times 10^{23}
C. 1.204×10241.204 \times 10^{24}
D. 2.408×10232.408 \times 10^{23}

✅ Correct Answer: A. 6.022×10236.022 \times 10^{23}

Explanation:

• 28 g N₂ = 1 mol.
• Molecules = 6.022×10236.022 \times 10^{23}.

Q97. What is the mass of 2.5 × 10²² molecules of CO₂?

A. 1.83 g
B. 0.183 g
C. 0.018 g
D. 0.25 g

✅ Correct Answer: B. 0.183 g

Explanation:

• Mole fraction = (2.5 × 10²²)/(6.022 × 10²³) = 0.0415 mol.
• Mass = 0.0415 × 44 ≈ 1.83 g.
  👉 Correct = A. 1.83 g.

Q98. The number of atoms in 2 g of H₂ is:

A. 6.022×10236.022 \times 10^{23}
B. 1.204×10241.204 \times 10^{24}
C. 3.011×10233.011 \times 10^{23}
D. 2.408×10232.408 \times 10^{23}

✅ Correct Answer: B. 1.204×10241.204 \times 10^{24}

Explanation:

• 2 g H₂ = 1 mol molecules = 6.022×10236.022 \times 10^{23}.
• Each molecule has 2 atoms → total = 1.204×10241.204 \times 10^{24}.

Q99. Which has greater number of molecules: 2 g H₂ or 32 g O₂?

A. 2 g H₂
B. 32 g O₂
C. Both equal
D. Cannot be compared

✅ Correct Answer: C. Both equal

Explanation:

• 2 g H₂ = 1 mol = Avogadro molecules.
• 32 g O₂ = 1 mol = Avogadro molecules.
• Both same.

Q100. Which of the following represents Avogadro’s hypothesis?

A. Equal volumes of gases at same T & P contain equal number of molecules.
B. Equal masses of gases contain equal number of atoms.
C. Equal volumes of gases at same T & P contain equal masses.
D. Equal moles of gases have unequal number of molecules.

✅ Correct Answer: A. Equal volumes of gases at same T & P contain equal number of molecules.

Explanation:

• A. Correct, Avogadro’s Law.
• B. Wrong, depends on atomic mass.
• C. Wrong, mass differs.
• **D. Wrong, 1 mole always = Avogadro’s number.

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