Question

3.92 g ferrous ammonium sulphate (FAS) consumes 50 ml of N/10 KMnO4. what is the percentage purity of the sample of FAS-
1)50% 2)78.4% 3)80% 4)39.2%
ans 1)50% pls ans urgently.

Answer 1

Answer: The correct option is 1.

Explanation: Normality of FAS is calculated by using the formula:

$Normality=\frac{\text{Weight of FAS}}{\text{Equivalent weight of FAS}\times Volume(\text{ in L})}$

Normality of 3.92 grams of FAS in 1L.

$Normality=\frac{3.92}{392\times 1}=0.01N$

Let percentage purity of FAS be 'x'

By using equation:

$N_1V_1=N_2V_2$

where,

$N_1\text{ and }V_1$ are the normality and volume of FAS

$N_2\text{ and }V_2$ are the normality and volume of $KMnO_4$

Putting value in above equation, we get:

$(\frac{x}{100})[0.01\times 1000]=[0.1\times 50]$

x = 50%