Correct option is A
To prepare 500 ml of 0.1N Na₂CO₃:
Normality (N) = 0.1
Volume (V) = 500 ml = 0.5 L
Equivalent weight of Na₂CO₃ = 106/2 = 53
Required mass = N × V × Eq. wt = 0.1 × 0.5 × 53 = 2.65 g
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Question
Quantity of Na₂CO₃ required to prepare 500 ml of 0.1N Na₂CO₃ is:
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