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Chemistry Practice Questions

NCERT Class IX Chapter 3: Atoms and Molecules

Created & Curated By S.K. Sinha

Section A: Questions
Q1.
The number of atoms present in 0.5 mole of nitrogen gas is:
(A) 3.01 × 1023
(B) 6.02 × 1023
(C) 1.20 × 1024
(D) 1.51 × 1023
Q2.
The molecular mass of H2SO4 is:
(Atomic masses: H = 1u, S = 32u, O = 16u)
(A) 96u
(B) 98u
(C) 94u
(D) 100u
Q3.
Which of the following contains maximum number of molecules?
(A) 1g of CO2
(B) 1g of N2
(C) 1g of H2
(D) 1g of CH4
Q4.
The formula unit mass of Ca3(PO4)2 is:
(Atomic masses: Ca = 40u, P = 31u, O = 16u)
(A) 310u
(B) 278u
(C) 354u
(D) 320u
Q5.
Which postulate of Dalton's atomic theory is the result of the law of conservation of mass?
(A) Atoms are indivisible
(B) Atoms can neither be created nor destroyed
(C) Atoms of the same element are identical
(D) Atoms combine in simple whole number ratios
Q6.
How many moles of CO2 are present in 88g of CO2?
(Atomic masses: C = 12u, O = 16u)
(A) 1 mole
(B) 2 moles
(C) 3 moles
(D) 0.5 mole
Q7.
The atomicity of phosphorus is:
(A) 1
(B) 2
(C) 3
(D) 4
Q8.
The mass of 0.2 mole of hydrogen atoms is:
(A) 0.2g
(B) 0.4g
(C) 2g
(D) 4g
Q9.
Which of the following represents Avogadro's number?
(A) 6.022 × 1022
(B) 6.022 × 1023
(C) 6.022 × 1024
(D) 6.022 × 1025
Q10.
What is the molecular formula of glucose if its empirical formula is CH2O and molecular mass is 180u?
(Atomic masses: C = 12u, H = 1u, O = 16u)
(A) C3H6O3
(B) C6H12O6
(C) C12H24O12
(D) CH2O
Q11.
The number of molecules in 1 mole of any gas at STP is:
(A) 6.022 × 1020
(B) 6.022 × 1021
(C) 6.022 × 1022
(D) 6.022 × 1023
Q12.
The mass of one mole of electrons is:
(Mass of electron = 9.1 × 10-31 kg)
(A) 9.1 × 10-31 kg
(B) 5.48 × 10-7 kg
(C) 1.67 × 10-27 kg
(D) 6.022 × 1023 kg
Q13.
If the molecular formula of a compound is C6H12O6, its empirical formula is:
(A) C6H12O6
(B) CH2O
(C) C2H4O2
(D) CHO
Q14.
According to the law of definite proportions, a chemical compound contains the constituent elements in:
(A) Different proportions by mass
(B) Fixed proportions by mass
(C) Different proportions by volume
(D) Simple ratios by volume
Q15.
The gram atomic mass of an element is numerically equal to:
(A) Atomic number
(B) Mass number
(C) Atomic mass
(D) Molecular mass
Q16.
How many atoms are present in a small piece of chalk containing 10g of calcium carbonate?
(Molecular mass of CaCO3 = 100u)
(A) 3.01 × 1022
(B) 6.02 × 1022
(C) 3.01 × 1023
(D) 6.02 × 1023
Q17.
The number of moles of H2O in 1.8g of water is:
(Molecular mass of H2O = 18u)
(A) 0.01 mole
(B) 0.1 mole
(C) 1 mole
(D) 10 moles
Q18.
The vapour density of a gas is 16. Its molecular mass will be:
(A) 16u
(B) 32u
(C) 8u
(D) 64u
Q19.
Which of the following has maximum number of atoms?
(A) 18g of H2O
(B) 18g of O2
(C) 18g of CO2
(D) 18g of CH4
Q20.
The law of multiple proportions is illustrated by:
(A) CO and CO2
(B) H2O and H2O2
(C) N2O and NO2
(D) All of the above
Q21.
Molar mass is the mass of:
(A) 1000 molecules of a substance
(B) 1 mole of a substance
(C) 100 molecules of a substance
(D) 1 molecule of a substance
Q22.
The atomicity of ozone is:
(A) 1
(B) 2
(C) 3
(D) 4
Q23.
The relative molecular mass of NH3 is:
(Atomic masses: N = 14u, H = 1u)
(A) 15u
(B) 16u
(C) 17u
(D) 18u
Q24.
2 moles of CO2 and 2 moles of H2O would have:
(A) Same number of molecules
(B) Same mass
(C) Same number of atoms
(D) Same volume
Q25.
The number of oxygen atoms in 0.1 mole of Na2CO3.10H2O is:
(A) 1.3 × 6.022 × 1023
(B) 0.13 × 6.022 × 1023
(C) 13 × 6.022 × 1023
(D) 0.013 × 6.022 × 1023
Section B: Answers
Question Answer
Q1 (B)
Q2 (B)
Q3 (C)
Q4 (A)
Q5 (B)
Q6 (B)
Q7 (D)
Q8 (A)
Q9 (B)
Q10 (B)
Q11 (D)
Q12 (B)
Q13 (B)
Q14 (B)
Q15 (C)
Q16 (C)
Q17 (B)
Q18 (B)
Q19 (D)
Q20 (D)
Q21 (B)
Q22 (C)
Q23 (C)
Q24 (A)
Q25 (A)
Section C: Solutions
Q1.
0.5 mol N2 = 0.5 × 6.022 × 1023 molecules = 3.01 × 1023 molecules; each has 2 atoms = 6.02 × 1023 atoms.
Q2.
Molecular mass = 2(1) + 32 + 4(16) = 2 + 32 + 64 = 98u.
Q3.
Moles = mass/molar mass; H2 has lowest molar mass (2g), hence maximum moles and molecules.
Q4.
Formula mass = 3(40) + 2(31) + 8(16) = 120 + 62 + 128 = 310u.
Q5.
Conservation of mass states atoms can neither be created nor destroyed in chemical reactions.
Q6.
Molar mass of CO2 = 44g; moles = 88/44 = 2 moles.
Q7.
Phosphorus exists as P4 molecules, so atomicity = 4.
Q8.
Mass = moles × atomic mass = 0.2 × 1 = 0.2g.
Q9.
Avogadro's number = 6.022 × 1023 particles per mole.
Q10.
Empirical formula mass = 30u; n = 180/30 = 6; molecular formula = C6H12O6.
Q11.
1 mole of any substance contains Avogadro's number of particles = 6.022 × 1023.
Q12.
Mass = 6.022 × 1023 × 9.1 × 10-31 = 5.48 × 10-7 kg.
Q13.
Simplest ratio: C:H:O = 6:12:6 = 1:2:1; empirical formula = CH2O.
Q14.
Law of definite proportions states elements combine in fixed mass ratios.
Q15.
Gram atomic mass equals atomic mass expressed in grams.
Q16.
Moles = 10/100 = 0.1; atoms in CaCO3 = 5; total = 0.1 × 5 × 6.022 × 1023 = 3.01 × 1023.
Q17.
Moles = 1.8/18 = 0.1 mole.
Q18.
Molecular mass = 2 × vapor density = 2 × 16 = 32u.
Q19.
18g CH4 = 18/16 = 1.125 mol; each has 5 atoms; maximum total atoms.
Q20.
All pairs show same elements combining in different ratios by mass.
Q21.
Molar mass = mass of 1 mole = 6.022 × 1023 particles.
Q22.
Ozone = O3; contains 3 oxygen atoms, atomicity = 3.
Q23.
Molecular mass = 14 + 3(1) = 17u.
Q24.
Equal moles contain equal number of molecules (Avogadro's law).
Q25.
Oxygen atoms: 3 (from CO3) + 10 (from 10H2O) = 13; total = 0.1 × 13 × 6.022 × 1023 = 1.3 × 6.022 × 1023.

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