Question

Find the normality of a solution prepared by dissolving 0.321 g of sodium carbonate $\left(N{a}_{2}C{O}_3\right)$ in 250 mL of water.

• A. 0.53 N
• B. 0.018 N
• C. 0.075 N
• D. 0.024 N

Answer 1

The correct option is D 0.024 N

Equivalent weight of $N{a}_{2}C{O}_3=\frac{\text{molar mass}}{\text{n-factor}}=\frac{106}{2}=53$

$\text{Normality}=\frac{\text{number of gram equivalents}}{\text{total volume of solution (mL)}}\times 1000=\frac{0.321}{53}\times \frac{1000}{250}=0.024N$