Question

$50c{m}^3$ of $0.04M{K}_{2}C{r}_2{O}_7$ in acidic medium oxidizes a sample of $H_{2}S$ gas to sulphur. Volume of $0.03M$ $KMn{O}_4$ required to oxidize the same amount of $H_{2}S$ gas to sulphur, in acidic medium is

• A. $80c{m}^3$
• B. $120c{m}^3$
• C. $60c{m}^3$
• D. $90c{m}^3$

Answer 1

Answer: (A)

$\because \frac{\text{ Normality }}{\text{ Molarity }}=\frac{\text{ Molecular weight }}{\text{ Equivalent weight }}$

For $K_{2}C{r}_2{O}_7$

$\therefore \frac{N_1}{0.04}=\frac{294}{49}$

$N_1=\frac{294}{49}\times 0.04=0.24$

For $KMn{O}_4,\frac{N_2}{0.03}=\frac{158}{31.6}$

$N_2=\frac{158}{31.6}\times 0.03=0.15$

Now, from normality equation

$N_1{V}_1=N_2{V}_2$

$\Rightarrow 0.24\times 50=0.15\times V_2$

$V_2=\frac{0.25\times 50}{0.15}=80c{m}^3$