Question

25 ml of 0.08 N Mohr's salt solution is oxidised by 20 ml of $K_{2}C{r}_2{O}_7$ in acidic medium. The mass of Mohr's salt present in 500 cc is:

• A. 3.96 g
• B. 15.7 g
• C. 39.6 g
• D. 39.2 g

Answer 1

The correct option is B 15.7 g

Mohr's salt is $FeS{O}_4\cdot (N{H}_4{)}_{2}S{O}_4\cdot 6H_{2}O$
25ml of 0.08N Mohr's salt solution corresponds to
$\frac{\text{25 ml}}{\text{1000 ml/L}}\times \text{0.08 geq/L}=\text{0.002 g eq}$
The equivalent weight of Mohr's salt is equal to its molecular weight as only on electron is involved in the oxidation
$F{e}^{2+}+e^{-}\to F{e}^{3+}$
The molecular weight of Mohr's salt is 392 g/mol.
The equivalent weight of Mohr's salt is 392 g/eq.
0.002 g eq of Mohr's salt corresponds to $\text{392 g /eq}\times \text{0.002 g eq}=\text{0.784 g}$
5 mL of solution contains 0.784 g of Mohr's salt.
500 mL of solution contains $\frac{\text{500 mL}}{\text{25 mL}}\times \text{0.784 g}=\text{15.7 g}$ of Mohr's salt.