Question

$3.92g/L$ of a sample of ferrous ammonium sulphate 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

Answer: (A)

$\underset{\text{(Ferrous ammonium sulphate) })}{N_1{V}_1}=\underset{\left(KMn{O}_4\right)}{N_2{V}_2}$
$N_1\times 1000=\frac{1}{10}\times 50$
$N_1=\frac{50}{10^4}=\frac{1}{200}$
Eq. wt. of $FeS{O}_4.{\left(N{H}_4\right)}_{2}S{O}_4.6H_{2}O$
$=$ mol. $wt.=392$
$\therefore$ Strength of pure salt
$=392\times \frac{1}{200}$
$=1.96g{L}^{-1}$
$\therefore \%$ purity
$=\frac{1.96}{3.92}\times 100=50\%$