CBSE Class 9 Chemistry Atoms and Molecules MCQs Set 05

Multiple Choice Questions (MCQs) for Class 9 Chemistry: Atoms and Molecules

Access targeted multiple-choice questions for Atoms and Molecules designed to align with the latest CBSE academic syllabus for Class 9 Chemistry. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

Practice Atoms and Molecules MCQs for Class 9 Chemistry

View or download the dedicated Atoms and Molecules MCQ resource below. Practicing these 50 objective questions regularly builds familiarity with standard exam patterns and helps secure higher marks in final Chemistry evaluations.

Multiple Choice Questions

Question 1: The chemical symbol of tungsten is
(a) W
(b) T
(c) Tu
(d) Tn
Show Answer & Explanation

Answer: (a) W

Explanation:
1. The symbol comes from its Latin or German name, Wolfram.

Teacher's Note:
1. Symbols can come from older names.

Question 2: The empirical formula of hydrogen peroxide (\( H_2O_2 \)) is
(a) \( H_2O_2 \)
(b) \( HO \)
(c) \( H_2O \)
(d) \( HO_2 \)
Show Answer & Explanation

Answer: (b) \( HO \)

Explanation:
1. The ratio H : O in \( H_2O_2 \) is \( 2:2=1:1 \).
2. The empirical formula is the simplest whole-number ratio, so it is \( HO \).

Teacher's Note:
1. Divide the subscripts by their highest common factor.

Question 3: If the formula of chloric acid is \( HClO_3 \), the formula of barium chlorate is
(a) \( BaClO_3 \)
(b) \( Ba_2(ClO_3) \)
(c) \( Ba(ClO_3)_2 \)
(d) \( Ba_2(ClO_3)_3 \)
Show Answer & Explanation

Answer: (c) \( Ba(ClO_3)_2 \)

Explanation:
1. The chlorate ion is \( ClO_3^- \) and barium is \( Ba^{2+} \).
2. Balancing charges needs two chlorate ions: \( Ba(ClO_3)_2 \).

Teacher's Note:
1. Criss-cross the charges.

Question 4: 0.6 mole of carbon dioxide weighs
(a) 44 g
(b) 7.3 g
(c) 264 g
(d) 26.4 g
Show Answer & Explanation

Answer: (d) 26.4 g

Explanation:
1. Molar mass of \( CO_2=12+32=44 \) g/mol.
2. Mass \( =0.6\times44=26.4 \) g.

Teacher's Note:
1. Mass = moles times molar mass.

Question 5: Sal ammoniac is the trade name of
(a) ammonium chloride
(b) ammonium hydroxide
(c) ammonium carbonate
(d) ammonium sulphate
Show Answer & Explanation

Answer: (a) ammonium chloride

Explanation:
1. Sal ammoniac is \( NH_4Cl \).

Teacher's Note:
1. Remember the common names.

Question 6: Epsom salt is
(a) sodium sulphate decahydrate
(b) hydrated magnesium sulphate, \( MgSO_4\cdot7H_2O \)
(c) calcium sulphate dihydrate
(d) hydrated copper sulphate
Show Answer & Explanation

Answer: (b) hydrated magnesium sulphate, \( MgSO_4\cdot7H_2O \)

Explanation:
1. Epsom salt is magnesium sulphate heptahydrate.

Teacher's Note:
1. Epsom salt contains magnesium.

Question 7: The balanced equation for the decomposition of potassium permanganate on heating is
(a) \( 2KMnO_4\to MnO_2+K_2O+O_2 \)
(b) \( KMnO_4\to MnO_2+O_2 \)
(c) \( 2KMnO_4\to K_2MnO_4+MnO_2+O_2 \)
(d) \( 2KMnO_4\to MnO_4+2MnO_2+O_2 \)
Show Answer & Explanation

Answer: (c) \( 2KMnO_4\to K_2MnO_4+MnO_2+O_2 \)

Explanation:
1. Count atoms in the first option: K 2, Mn 2, O 8 on the left and K 2, Mn 2, O \( 4+2+2=8 \) on the right, so it is balanced.
2. The others are unbalanced or have wrong formulae.

Teacher's Note:
1. Check each element on both sides.

Question 8: The volume of oxygen at STP required for complete combustion of 1 mole of methane is
(a) 22.4 litres
(b) 11.2 litres
(c) 1.12 litres
(d) 44.8 litres
Show Answer & Explanation

Answer: (d) 44.8 litres

Explanation:
1. \( CH_4+2O_2\to CO_2+2H_2O \): 1 mole of methane needs 2 moles of oxygen.
2. \( 2\times22.4=44.8 \) litres.

Teacher's Note:
1. One mole of gas occupies 22.4 litres at STP.

Question 9: A solution of 2 g of NaCl reacts with 5 g of \( AgNO_3 \) in water. The maximum mass of AgCl formed is
(a) 4.22 g
(b) 1.80 g
(c) 3.65 g
(d) 8.2 g
Show Answer & Explanation

Answer: (a) 4.22 g

Explanation:
1. \( NaCl+AgNO_3\to AgCl+NaNO_3 \). Moles: NaCl \( =\frac{2}{58.5}=0.0342 \), \( AgNO_3=\frac5{170}=0.0294 \).
2. \( AgNO_3 \) is limiting: AgCl \( =0.0294\times143.5\approx4.22 \) g.

Teacher's Note:
1. Find the limiting reactant.

Question 10: 2 g of hydrogen react with 0.4 mole of chlorine to form HCl. The number of hydrogen molecules left behind is
(a) \( 4.0\times10^{23} \)
(b) \( 3.614\times10^{23} \)
(c) \( 6.023\times10^{23} \)
(d) \( 2.414\times10^{23} \)
Show Answer & Explanation

Answer: (b) \( 3.614\times10^{23} \)

Explanation:
1. 2 g of hydrogen is 1 mole. \( H_2+Cl_2\to2HCl \) uses 0.4 mole of hydrogen.
2. Left: \( 0.6 \) mole \( =0.6\times6.023\times10^{23}=3.614\times10^{23} \) molecules.

Teacher's Note:
1. Use moles.

Question 11: 2 litres of hydrogen at STP are passed over 10 g of heated cupric oxide (\( CuO=79.54 \)). The mass of cupric oxide remaining is
(a) 5.65 g
(b) 2.02 g
(c) 2.90 g
(d) 1.86 g
Show Answer & Explanation

Answer: (c) 2.90 g

Explanation:
1. \( CuO+H_2\to Cu+H_2O \): 22.4 litres of hydrogen react with 79.54 g of CuO.
2. 2 litres react with \( \frac{79.54\times2}{22.4}=7.10 \) g, so left \( =10-7.10=2.90 \) g.

Teacher's Note:
1. Use the 22.4 litre molar volume.

Question 12: 1 g of a mixture of sodium carbonate and sodium bicarbonate gives 0.816 g of residue on strong heating. The percentage of sodium bicarbonate in the mixture is
(a) 5 percent
(b) 10 percent
(c) 80 percent
(d) about 50 percent
Show Answer & Explanation

Answer: (d) about 50 percent

Explanation:
1. \( 2NaHCO_3\to Na_2CO_3+H_2O+CO_2 \): 168 g loses 62 g.
2. Loss \( =1-0.816=0.184 \) g, so bicarbonate \( =\frac{0.184\times168}{62}=0.4986 \) g, about 50 percent.

Teacher's Note:
1. Only the bicarbonate loses mass.

Question 13: The mass of hydrogen liberated at STP when 9.2 g of sodium reacts with water is
(a) 0.4 g
(b) 4 g
(c) 2.6 g
(d) 0.92 g
Show Answer & Explanation

Answer: (a) 0.4 g

Explanation:
1. \( 2Na+2H_2O\to2NaOH+H_2 \): 46 g of sodium gives 2 g of hydrogen.
2. \( \frac{9.2\times2}{46}=0.4 \) g.

Teacher's Note:
1. Use the mole ratio 2 : 1.

Question 14: The volume of oxygen at STP obtained when 7.8 g of sodium peroxide reacts with water is
(a) 11.2 litres
(b) 1.12 litres
(c) 2.24 litres
(d) 44.8 litres
Show Answer & Explanation

Answer: (b) 1.12 litres

Explanation:
1. \( 2Na_2O_2+2H_2O\to4NaOH+O_2 \): 156 g gives 22.4 litres of oxygen.
2. \( \frac{7.8\times22.4}{156}=1.12 \) litres.

Teacher's Note:
1. Moles of peroxide are 0.1.

Question 15: The volume of oxygen at STP required to burn 1 g of glucose completely is
(a) 1.12 litres
(b) 2.08 litres
(c) 0.747 litres
(d) 1.94 litres
Show Answer & Explanation

Answer: (c) 0.747 litres

Explanation:
1. \( C_6H_{12}O_6+6O_2\to6CO_2+6H_2O \): 180 g needs \( 6\times22.4 \) litres.
2. \( \frac{1\times134.4}{180}=0.747 \) litres.

Teacher's Note:
1. Use the equation.

Question 16: 250 ml of a gas at STP weighs 0.491 g. The molecular mass of the gas is
(a) 22
(b) 64
(c) 71
(d) 44
Show Answer & Explanation

Answer: (d) 44

Explanation:
1. Mass of 22.4 litres \( =\frac{0.491\times22400}{250}=43.99 \) g.
2. So the molecular mass is about 44.

Teacher's Note:
1. Mass of one molar volume.

Question 17: The mass of sulphur which, on heating with concentrated nitric acid, forms 100 g of sulphuric acid is about
(a) 32.65 g
(b) 3.6 g
(c) 18.4 g
(d) 12.96 g
Show Answer & Explanation

Answer: (a) 32.65 g

Explanation:
1. \( S\to H_2SO_4 \): 32 g of sulphur gives 98 g of acid.
2. \( \frac{100\times32}{98}=32.65 \) g.

Teacher's Note:
1. One atom of sulphur gives one molecule of acid.

Question 18: The volume of oxygen at 20 degrees Celsius and 720 mm pressure needed to burn 2.2 g of propane completely is about
(a) 5.6 litres
(b) 6.35 litres
(c) 5.74 litres
(d) 9.8 litres
Show Answer & Explanation

Answer: (b) 6.35 litres

Explanation:
1. \( C_3H_8+5O_2\to3CO_2+4H_2O \): 44 g needs \( 5\times22.4 \) litres at STP, so 2.2 g needs 5.6 litres at STP.
2. \( V=5.6\times\frac{293}{273}\times\frac{760}{720}\approx6.35 \) litres.

Teacher's Note:
1. Convert the STP volume with the gas law.

Question 19: The smallest possible molecular mass of any substance is 2, which is the molecular mass of
(a) helium
(b) a hydrogen atom
(c) hydrogen (\( H_2 \))
(d) water
Show Answer & Explanation

Answer: (c) hydrogen (\( H_2 \))

Explanation:
1. Hydrogen is the lightest element and exists as \( H_2 \) with mass \( 2 \).
2. No molecule can be lighter.

Teacher's Note:
1. Lightest element.

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Multiple Choice Questions (MCQs) for Class 9 Chemistry Atoms and Molecules

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FAQs

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