Question

$5g$ sample contain only $N{a}_{2}C{O}_3$ and $N{a}_{2}S{O}_4$. This sample is dissolved and the volume made up to $250$ mL. $25$ mL of this solution neutralizes $20$ mL of $0.1M{H}_{2}S{O}_4$. Calculate the $\%$ of $N{a}_{2}S{O}_4$ in the sample.

• A. $42.4$
• B. $57.6$
• C. $36.2$
• D. None of the above

Answer 1

The correct option is B $57.6$

Only $N{a}_{2}C{O}_3$ is react with $H_{2}S{O}_4$.
$N{a}_{2}C{O}_3+H_{2}S{O}_4\to N{a}_{2}S{O}_4+H_{2}C{O}_3$
Milli-moles of $N{a}_{2}C{O}_3$ $=$ Milli-moles of $H_{2}S{O}_4=20\times 0.1=2$
Milli-moles of $N{a}_{2}C{O}_3$ in $250$ mL of solution$=\frac{250}{25}\times 2=20$
Weight of $N{a}_{2}C{O}_3=20\times 106\times {10}^{-3}=2.12$ g
$\%$ of $N{a}_{2}C{O}_3=\frac{2.12}{5}\times 100=42.4$
$\therefore \%$ of $N{a}_{2}S{O}_4=100-42.4=57.6\%$