Question

$4488mL$ of $S{O}_2$ at $NTP$ is passed through $100mL$ of a $0.2N$ solution of $NaOH$. Find the weight of the salt formed.

Answer 1

$4488mL$ of $S{O}_2\underrightarrow{NTP}$ $100mL$ of $0.2N$ $NaOH$

miliequivalent of $S{O}_2=\frac{4488}{2244}$

$20$

miliequivalent of $NaOH=0.2\times 100$

$=20m-eq$

Hence, miliequivalent of $N{a}_{2}S{O}_3$ formed $=20$

mass=$\frac{20}{1000}\times 53=\frac{53}{50}$

$=0.106g$