Quantity of Na₂CO₃ required to prepare 500 ml of 0.1 N Na₂CO₃ is ____.

Correct option is A

Correct option: (a) 2.65 g
Step-by-step explanation:
· Normality (N) = equivalents per litre
· Equivalent weight of Na₂CO₃:
· Molecular weight = 106
· Valency factor = 2
· Equivalent weight = 106 / 2 = 53
· Formula used: Weight [g] = Normality × Equivalent weight × Volume [L]
· Substituting values:
· = 0.1 × 53 × 0.5
· = 2.65 g
➡️ Hence, 2.65 g of Na₂CO₃ is required.
❌ Why other options are incorrect
· (b) 0.265 g
· One decimal place error.
· (c) 26.5 g
· Ten times higher than required.
· (d) 0.00265 g
· Incorrect unit scale (milligram range).
�� Key Points
· For salt solutions, always use equivalent weight.
· Na₂CO₃ is dibasic, hence valency factor = 2.
· Volume must be converted to litres.
✔️ Final Answer: (a)