To prepare \(250\, \ mL\) standard solution of \(M/20\, \ NaCl\), amount of \(NaCl\) required is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 0.73 g
Moles of NaCl needed = molarity × volume in litres:
n = (1/20) mol L−1 × 0.250 L = 0.0125 mol
Molar mass of NaCl = 23 + 35.5 = 58.5 g mol−1
Mass = 0.0125 mol × 58.5 g mol−1 = 0.731 g ≈ 0.73 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.”
Note: The same/similar question has been asked in different years by NTA as per the details given below:
| 2026 | 2025 | 2024 | 2023 | 2022 |
|---|---|---|---|---|
| – | – | – | 28-05 S1 24-05 S3 | – |