CBSE Class 9 Chemistry Atoms and Molecules MCQs Set 04

Download CBSE MCQs for Class 9 Chemistry: Atoms and Molecules

Review structured MCQ sets for Class 9 Chemistry Atoms and Molecules. Built according to official CBSE guidelines, these downloadable questions support daily revision and core concept reinforcement.

Chapter-wise Objective Questions: Atoms and Molecules

Navigate directly to the 50 objective questions for Atoms and Molecules using the digital viewer below. Each practice set includes verified answer keys, allowing students to instantly cross-check their work and identify areas requiring further revision.

Multiple Choice Questions

Question 1: Na\( _2 \)C\( _2 \)O\( _4 \) is the formula of sodium
(a) carbonate
(b) formate
(c) oxalate
(d) acetate
Show Answer & Explanation

Answer: (c) oxalate

Explanation:
1. The C2O4 2− ion is oxalate.
2. Sodium oxalate is Na2C2O4.

Teacher's Note:
1. Carbonate is CO3 2−; formate is HCOO−; acetate is CH3COO−.

Question 2: 12 g of Mg (at. wt. 24) will react completely with an acid to give
(a) one mole of H\( _2 \)
(b) one-half mole of H\( _2 \)
(c) one mole of O\( _2 \)
(d) None of the above
Show Answer & Explanation

Answer: (b) one-half mole of H\( _2 \)

Explanation:
1. Mg + 2H+ → Mg2+ + H2.
2. 12 g Mg = 0.5 mol, which gives 0.5 mol of H2.

Teacher's Note:
1. One mole of Mg gives one mole of H2.
2. 12 g is half a mole of magnesium.

Question 3: An Avogadro number of molecules of hydrogen weigh
(a) half gram molecule
(b) one gram atom
(c) 2 g
(d) 1 g
Show Answer & Explanation

Answer: (c) 2 g

Explanation:
1. One mole of H2 molecules weighs 2 g.
2. So Avogadro's number of hydrogen molecules weighs 2 g.

Teacher's Note:
1. H2 has molar mass 2 g.

Question 4: The number of moles of CO that contain the same number of gram atoms as present in 4 moles of NH\( _3 \) is
(a) 4 moles
(b) 2 moles
(c) 6 moles
(d) 8 moles
Show Answer & Explanation

Answer: (d) 8 moles

Explanation:
1. 4 moles of NH3 contain 4 × 4 = 16 gram atoms.
2. Each mole of CO has 2 gram atoms, so 16/2 = 8 moles are needed.

Teacher's Note:
1. NH3 has 4 atoms per molecule; CO has 2.

Question 5: A formula tells about
(a) the number of atoms present in one molecule
(b) a mathematical formula
(c) an element only
(d) the source of the compound
Show Answer & Explanation

Answer: (a) the number of atoms present in one molecule

Explanation:
1. A chemical formula shows the elements present and the number of atoms of each in one molecule or formula unit.

Teacher's Note:
1. The subscripts give the number of atoms.

Question 6: Weight of one gram atom is
(a) the atomic weight expressed in grams
(b) the weight in grams of one atom
(c) atoms present in 1 g
(d) none of these
Show Answer & Explanation

Answer: (a) the atomic weight expressed in grams

Explanation:
1. One gram atom is one mole of atoms.
2. Its weight is the atomic weight expressed in grams.

Teacher's Note:
1. Gram atom means atomic mass in grams.

Question 7: The law of multiple proportions can be applied to which pair of compounds out of CO\( _2 \), SO\( _2 \), SO\( _3 \), H\( _2 \)O?
(a) CO\( _2 \), SO\( _2 \)
(b) CO\( _2 \), SO\( _3 \)
(c) SO\( _3 \), H\( _2 \)O
(d) SO\( _2 \), SO\( _3 \)
Show Answer & Explanation

Answer: (d) SO\( _2 \), SO\( _3 \)

Explanation:
1. The law applies to two compounds formed by the same two elements.
2. SO2 and SO3 are both made of sulphur and oxygen, with oxygen masses in the ratio 2 : 3 for a fixed mass of sulphur.

Teacher's Note:
1. Look for the same pair of elements in both compounds.

Question 8: The solubility of a salt at 50°C is 16 g per 100 g of water. The least weight of water needed to dissolve 4 g of this substance at 50°C is
(a) 100 g
(b) 250 g
(c) 25 g
(d) 16 g
Show Answer & Explanation

Answer: (c) 25 g

Explanation:
1. 16 g dissolves in 100 g of water.
2. 4 g needs 100 × 4/16 = 25 g of water.

Teacher's Note:
1. Set up a direct proportion between solute and water.
2. Divide first, then multiply.

Question 9: 2.76 g of silver carbonate on being strongly heated yields a residue weighing
(a) 2.48 g
(b) 2.16 g
(c) 2.32 g
(d) 2.64 g
Show Answer & Explanation

Answer: (b) 2.16 g

Explanation:
1. Ag2CO3 → 2Ag + CO2 + ½O2 on strong heating.
2. 2.76 g Ag2CO3 = 2.76/276 = 0.01 mol, which gives 0.02 mol of Ag.
3. Residue = 0.02 × 108 = 2.16 g.

Teacher's Note:
1. On strong heating silver carbonate leaves metallic silver.
2. Ag2CO3 has molar mass 276.

Question 10: The formula of ammonium ferricyanide is
(a) Am\( _3 \)[Fe(CN)\( _6 \)]
(b) NH\( _4 \)[Fe(CN)\( _6 \)]
(c) (NH\( _4 \))\( _3 \)[Fe(CN)\( _6 \)]
(d) NH\( _4 \)[Fe(CN)\( _6 \)]\( _3^- \)
Show Answer & Explanation

Answer: (c) (NH\( _4 \))\( _3 \)[Fe(CN)\( _6 \)]

Explanation:
1. Ferricyanide is [Fe(CN)6]3−, with charge 3−.
2. Three ammonium ions (NH4)+ balance it, giving (NH4)3[Fe(CN)6].

Teacher's Note:
1. Balance the charges of the cation and the anion.
2. The ammonium ion is NH4+.

Question 11: 100 mL of PH\( _3 \) is decomposed into phosphorus and hydrogen. The change in volume is
(a) 50 mL increase
(b) 500 mL decrease
(c) 900 mL decrease
(d) Nil
Show Answer & Explanation

Answer: (a) 50 mL increase

Explanation:
1. 4PH3 → P4 (solid) + 6H2.
2. 4 volumes of PH3 give 6 volumes of H2 (phosphorus is solid).
3. 100 mL PH3 gives 150 mL H2, so the volume increases by 50 mL.

Teacher's Note:
1. Solid products do not count in gas volumes.
2. Use the gas volume ratio from the balanced equation.

Question 12: The kinetic energy of a gas molecule would be zero at
(a) 0°C
(b) 273°C
(c) −273°C
(d) 116°C
Show Answer & Explanation

Answer: (c) −273°C

Explanation:
1. Kinetic energy is proportional to the absolute temperature.
2. It becomes zero at 0 K, which is about −273°C.

Teacher's Note:
1. Absolute zero is 0 K = −273°C.

Question 13: Which of the following phrases would be incorrect to use?
(a) a molecule of an element
(b) an atom of an element
(c) a molecule of a compound
(d) an atom of a compound
Show Answer & Explanation

Answer: (d) an atom of a compound

Explanation:
1. A compound is made of molecules or formula units, which in turn contain atoms of the elements.
2. An atom belongs to an element, so 'an atom of a compound' is incorrect.

Teacher's Note:
1. Atoms belong to elements; molecules belong to elements or compounds.

Question 14: Six gram of magnesium (atomic weight 24) reacts with excess of an acid. The amount of hydrogen produced would be
(a) 0.5 g
(b) 1.0 g
(c) 2 g
(d) 4 g
Show Answer & Explanation

Answer: (a) 0.5 g

Explanation:
1. 6 g Mg = 0.25 mol, giving 0.25 mol H2.
2. Mass of H2 = 0.25 × 2 = 0.5 g.

Teacher's Note:
1. Moles of Mg equal moles of H2.
2. Convert moles of H2 to grams at the end.

Question 15: The correct increasing order of molecular weights is
(a) H\( _2 \)O > H\( _2 \)S > CO\( _2 \) > SO\( _2 \)
(b) H\( _2 \)O > H\( _2 \)S < CO\( _2 \) < SO\( _2 \)
(c) H\( _2 \)O < H\( _2 \)S < CO\( _2 \) < SO\( _2 \)
(d) H\( _2 \)O > H\( _2 \)S < CO\( _2 \) > SO\( _2 \)
Show Answer & Explanation

Answer: (c) H\( _2 \)O < H\( _2 \)S < CO\( _2 \) < SO\( _2 \)

Explanation:
1. Molecular weights: H2O = 18, H2S = 34, CO2 = 44, SO2 = 64.
2. The increasing order is H2O < H2S < CO2 < SO2.

Teacher's Note:
1. Compute each molecular weight first.
2. Check the direction of the signs.

Question 16: For liquefaction of a gas the favourable conditions are
(a) low temperature and pressure
(b) low temperature and catalyst
(c) low temperature and high pressure
(d) low temperature, high pressure and a catalyst
Show Answer & Explanation

Answer: (c) low temperature and high pressure

Explanation:
1. Lowering the temperature reduces kinetic energy of molecules.
2. High pressure brings the molecules close together so that attractions hold them as a liquid.
3. A catalyst has no role.

Teacher's Note:
1. Cooling and compressing liquefy a gas.

Question 17: The ion that is not univalent is
(a) ammonium
(b) sodium
(c) bicarbonate
(d) sulphite
Show Answer & Explanation

Answer: (d) sulphite

Explanation:
1. Ammonium (NH4+), sodium (Na+) and bicarbonate (HCO3−) all carry a single charge.
2. Sulphite (SO3 2−) carries two charges, so it is divalent.

Teacher's Note:
1. Univalent means a charge of one.

Question 18: 56 g of CaO is mixed with 63 g of HNO\( _3 \). The amount of Ca(NO\( _3 \))\( _2 \) formed is
(a) 4 g
(b) 3.28 g
(c) 164 g
(d) 82 g
Show Answer & Explanation

Answer: (d) 82 g

Explanation:
1. CaO + 2HNO3 → Ca(NO3)2 + H2O.
2. 56 g CaO = 1 mol; 63 g HNO3 = 1 mol.
3. HNO3 is limiting: 1 mol gives 0.5 mol Ca(NO3)2 = 0.5 × 164 = 82 g.

Teacher's Note:
1. Find the limiting reagent before calculating the product.
2. Two moles of HNO3 are needed per mole of CaO.

Question 19: How much quicklime can be obtained from 25 g of CaCO\( _3 \)?
(a) 28 g
(b) 14 g
(c) 56 g
(d) None of these
Show Answer & Explanation

Answer: (b) 14 g

Explanation:
1. 100 g CaCO3 gives 56 g CaO.
2. 25 g gives 25 × 56/100 = 14 g.

Teacher's Note:
1. Use the mass ratio 100 : 56.
2. One quarter of 100 g gives one quarter of 56 g.

Question 20: A balanced chemical equation supports the
(a) law of chemical equilibrium
(b) law of conservation of energy
(c) law of conservation of mass
(d) none of these
Show Answer & Explanation

Answer: (c) law of conservation of mass

Explanation:
1. A balanced equation has equal numbers of each kind of atom on both sides.
2. This supports the law of conservation of mass.

Teacher's Note:
1. Balancing is done to keep mass the same.

Atoms and Molecules Objective Questions & Solutions for Class 9 Chemistry

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

Test your conceptual understanding of Atoms and Molecules with these targeted multiple-choice questions. Designed in alignment with the latest CBSE curriculum for Class 9 Chemistry, these problem sets build accuracy and prepare students for objective exams.

Concept Clarification for Atoms and Molecules

Built using the official NCERT book for Class 9, these Chemistry objective sets provide reliable academic guidance. Pair your practice with our recommended NCERT solutions to master optimal problem-solving approaches.

Additional Study Resources for Class 9 Chemistry

Wrap up your chapter revision by testing your knowledge against standard question formats. Everything on our platform is provided free of charge.

FAQs

Where can I access latest CBSE Class 9 Chemistry Atoms and Molecules MCQs Set 04?

You can get most exhaustive CBSE Class 9 Chemistry Atoms and Molecules MCQs Set 04 for free on StudiesToday.com. These MCQs for Class 9 Chemistry are updated for the 2026-27 academic session as per CBSE examination standards.

Are Assertion-Reasoning and Case-Study MCQs included in the Chemistry Class 9 material?

Yes, our CBSE Class 9 Chemistry Atoms and Molecules MCQs Set 04 include the latest type of questions, such as Assertion-Reasoning and Case-based MCQs. 50% of the CBSE paper is now competency-based.

How do practicing Chemistry MCQs help in scoring full marks in Class 9 exams?

By solving our CBSE Class 9 Chemistry Atoms and Molecules MCQs Set 04, Class 9 students can improve their accuracy and speed which is important as objective questions provide a chance to secure 100% marks in the Chemistry.

Do you provide answers and explanations for CBSE Class 9 Chemistry Atoms and Molecules MCQs Set 04?

Yes, Chemistry MCQs for Class 9 have answer key and brief explanations to help students understand logic behind the correct option as its important for 2026 competency-focused CBSE exams.

Can I practice these Chemistry Class 9 MCQs online?

Yes, you can also access online interactive tests for CBSE Class 9 Chemistry Atoms and Molecules MCQs Set 04 on StudiesToday.com as they provide instant answers and score to help you track your progress in Chemistry.