Question

5.3 g of $M_{2}C{O}_3$ is dissolved in 150 mL of 1 N $HCl$. Unused acid required 100 mL of 0.5 N $NaOH$. Hence, equivalent weight of M is :

• A. 23
• B. 12
• C. 24
• D. 13

Answer 1

Answer: (A)

23

100 mL of 0.5 N NaOH will neutralize 50 mL of 1 N HCl.
The volume of 1 N HCl that reacts with metal carbonate is $150-50=100mL$
This corresponds to 0.1 equivalent of HCl.
This is also equal to 0.1 equivalent of metal carbonate.
The equivalent mass of metal carbonate is $\frac{2M+60}{2}=(M+30)$
The number of equivalents is $0.1=\frac{5.3}{M+30}$.
Hence, $M=23$