Question

$3.92g/L$ of a sample of ferrous ammonium sulphate $(FeS{O}_4.(N{H}_4{)}_{2}S{O}_4.6H_{2}O)$ reacts completely with $50mL\frac{N}{10}KMn{O}_4$ solution. The percentage purity of the sample is :

• A. $50\%$
• B. $78.4\%$
• C. $80.0\%$
• D. $39.2\%$

Answer 1

The correct option is A $50\%$

$\text{Ferrous ammonium sulphate}\equiv FeS{O}_4.(N{H}_4{)}_{2}S{O}_4.6H_{2}O$
$\text{Equivalents of}FeS{O}_4.(N{H}_4{)}_{2}S{O}_4.6H_{2}O=\text{Equivalents of}KMn{O}_4$
$N_1{V}_1=N_2{V}_2$
$N_1\times 1000=\frac{1}{10}\times 50$
$N_1=\frac{1}{200}$
$\text{Equivalent weight of}FeS{O}_4.(N{H}_4{)}_{2}S{O}_4.6H_{2}O=\text{Molecular weight}\times 1=392$

$\therefore \text{Strength of pure salt}=392\times \frac{1}{200}=1.96g/L$
$\therefore$ % purity $=\frac{1.96}{3.92}\times 100=50\%$