Some Basic Concepts of Chemistry MCQs (Difficulty-wise) with Answers
This difficulty-wise set helps you revise Some Basic Concepts of Chemistry in the right order: mole and particles first, then formulas and composition, then stoichiometry and limiting reagent. Use the explanations to fix the step where you made the mistake.
Easy (Level 1)
Question 1
One mole of any substance contains:
A. $6.022\times 10^{23}$ particles
B. $6.022\times 10^{23}$ grams
C. $22.4$ particles
D. $22.4$ grams
Correct Answer: A
Explanation: One mole contains Avogadro’s number of entities.
Question 2
Number of moles in 18 g of water ($H_2O$) is:
A. 0.5
B. 1
C. 2
D. 18
Correct Answer: B
Explanation: Molar mass of $H_2O$ is 18 g/mol, so 18 g = 1 mol.
Question 3
Molar mass of $Na_2CO_3$ is:
A. 84 g/mol
B. 96 g/mol
C. 106 g/mol
D. 116 g/mol
Correct Answer: C
Explanation: $2\times 23 + 12 + 3\times 16 = 106,g,mol^{-1}$.
Medium (Level 2)
Revision Tip: For empirical/molecular formula, calculate empirical formula mass first, then use factor $= \frac{\text{molar mass}}{\text{empirical mass}}$.
Question 4
A compound has 40% C, 6.67% H, and 53.33% O by mass. The empirical formula is:
A. $CH_2O$
B. $C_2H_4O_2$
C. $CH_4O$
D. $C_2H_6O$
Correct Answer: A
Explanation: Take 100 g: C = 40/12 ≈ 3.33, H = 6.67/1 = 6.67, O = 53.33/16 ≈ 3.33. Divide by 3.33 → 1 : 2 : 1 → $CH_2O$.
Question 5
If a compound has empirical formula $CH_2$ and molar mass 56 g/mol, its molecular formula is:
A. $C_2H_4$
B. $C_3H_6$
C. $C_4H_8$
D. $C_5H_{10}$
Correct Answer: C
Explanation: Empirical mass = 14. Factor $= 56/14 = 4$. Molecular formula $= (CH_2)_4 = C_4H_8$.
Question 6
Molality (m) is defined as:
A. Moles of solute per litre of solution
B. Moles of solute per kg of solvent
C. Grams of solute per kg of solution
D. Moles of solute per litre of solvent
Correct Answer: B
Explanation: Molality uses mass of solvent (kg), not volume.
Hard (Level 3)
Revision Tip: In limiting reagent problems, compare “available moles / stoichiometric coefficient” for each reactant.
Question 7
For the reaction $N_2 + 3H_2 \rightarrow 2NH_3$, if 1 mol $N_2$ and 2 mol $H_2$ are taken, the limiting reagent is:
A. $N_2$
B. $H_2$
C. Both are limiting
D. None is limiting
Correct Answer: B
Explanation: 1 mol $N_2$ needs 3 mol $H_2$. Only 2 mol available, so $H_2$ limits.
Question 8
In the reaction $2Al + 3Cl_2 \rightarrow 2AlCl_3$, if 1 mol Al reacts with 1 mol $Cl_2$, the limiting reagent is:
A. Al
B. $Cl_2$
C. Both
D. Cannot be determined
Correct Answer: B
Explanation: For 1 mol Al, required $Cl_2 = \frac{3}{2} = 1.5$ mol. Only 1 mol $Cl_2$ available.
Question 9
If 0.2 mol of a solute is dissolved to make 500 mL solution, the molarity is:
A. 0.1 M
B. 0.2 M
C. 0.4 M
D. 2.5 M
Correct Answer: C
Explanation: Volume = 0.5 L. $M = \frac{0.2}{0.5} = 0.4$ M.
Question 10
The number of atoms in 0.25 mol of helium is:
A. $1.5055\times 10^{23}$
B. $3.011\times 10^{23}$
C. $6.022\times 10^{23}$
D. $1.2044\times 10^{24}$
Correct Answer: A
Explanation: Atoms $= nN_A = 0.25\times 6.022\times 10^{23} = 1.5055\times 10^{23}$.
Want a mixed set for quick revision? Go back to: Some Basic Concepts of Chemistry MCQs (Mixed).
If you’re short on time, do Level 1 + Level 2 fully, then attempt only 2–3 Level 3 questions.