Question

$11.65\text{g}$ of $BaS{O}_4$ was obtained as a dry precipitate when $10\text{g}$ of an impure sample of $N{a}_{2}S{O}_4$ was dissolved in water and treated with an excess of $BaC{l}_2$. Calculate the percentage purity of the $N{a}_{2}S{O}_4$ sample. (Molar mass of $BaS{O}_4=233\text{g}andN{a}_{2}SO4=142\text{g}$)

• A. 38 %
• B. 71 %
• C. 22 %
• D. 88 %

Answer 1

The correct option is B 71 %

$N{a}_{2}S{O}_4+BaC{l}_2\to BaS{O}_4\downarrow +2NaCl$
$\text{Moles of}BaS{O}_4=\frac{\text{given mass}}{\text{molar mass}}=\frac{11.65}{233}=0.05\text{mol}$
According to the reaction, $1$ mole of $BaS{O}_4$ precipitate is obtained from $1$ mole of $N{a}_{2}S{O}_4$
$0.05$ moles of $BaS{O}_4$ will be produced by $0.05$ mole of $N{a}_{2}S{O}_4$
Amount of pure $N{a}_{2}S{O}_4$ required = $0.05$ mol $\times 142$ $\left(\frac{\text{g}}{\text{mol}}\right)=7.1\text{g}$
$\text{Percentage purity}=\frac{\text{mass of pure compound in sample}}{\text{total mass of impure sample}}\times 100=\frac{7.1}{10}\times 100$
$=71\%$