Question

$5.3\, g$ of $M_2 CO _3$ 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)

$V\, mL$ of $1 N$ unused $HCl =100\, mL$ of $0.5 \,N\, NaOH$
$V=50\, mL$
Used $1 \,N\,HCl =100\, mL$
$100 \,mL$ of $1 \,N \,HCl =0.1$ equivalent $HCl$
$=0.1$ equivalent $HCl$
$=0.11$ equivalent of $M_2 CO _3$
$\therefore$ equivalent mass of $M_2 CO _3=\frac{2 M+60}{2}=(M+30)$
$\therefore \frac{5.3}{M+30}=0.1$
$\therefore M=23$