Match List I with List II. For preparing \(250 mL\) of aqueous solution with molarity \(0.1\) (List I) the amount of salt required (List II) is:
| List I | List II |
| A. \(0.1 M\, \ NaOH\) | I. 1.46 g |
| B. \(0.1 M\, \ HCl\) | II. 1.4 g |
| C. \(0.1 M\, \ NaCl\) | III. 1 g |
| D. \(0.1 M\, \ KOH\) | IV. 0.9 g |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → A-III, B-IV, C-I, D-II
Molarity is the number of moles of solute in one litre of solution. The mass needed for a given volume is therefore
\( \displaystyle \text{mass} = M \times V(\text{L}) \times \text{molar mass} \)
Every solution here is 0.1 M and 250 mL, so the moles needed are the same for all four:
\( \displaystyle n = 0.1\ \text{mol L}^{-1} \times 0.250\ \text{L} = 0.025\ \text{mol} \)
A. 0.1 M NaOH — This matches III. 1 g. Molar mass 23 + 16 + 1 = 40 g mol−1; 0.025 × 40 = 1.0 g.
B. 0.1 M HCl — This matches IV. 0.9 g. Molar mass 1 + 35.5 = 36.5 g mol−1; 0.025 × 36.5 = 0.91 g ≈ 0.9 g.
C. 0.1 M NaCl — This matches I. 1.46 g. Molar mass 23 + 35.5 = 58.5 g mol−1; 0.025 × 58.5 = 1.46 g.
D. 0.1 M KOH — This matches II. 1.4 g. Molar mass 39 + 16 + 1 = 56 g mol−1; 0.025 × 56 = 1.4 g.
Shortcut: the moles are equal, so the masses follow the molar masses: HCl (36.5) < NaOH (40) < KOH (56) < NaCl (58.5) pairs with 0.9 < 1 < 1.4 < 1.46 g.
NCERT (2026-27 edition):
Molarity – “Molarity (M) is defined as number of moles of solute dissolved in one litre (or one cubic decimetre) of solution,
\( \text{Molarity} = \frac{\text{Moles of solute}}{\text{Volume of solution in litre}} \)”