Question

$50\,cm^3$ of $0.04 \,M \,K_2 Cr_2 O_7$ in acidic medium oxidizes a sample of $H_2S$ gas to sulphur. Volume of $0.03\, M$ $KMnO_4$ required to oxidize the same amount of $H_2S$ gas to sulphur, in acidic medium is

• A. $80\,cm^3$
• B. $120\,cm^3$
• C. $60\,cm^3$
• D. $90\,cm^3$

Answer 1

Answer: (A)

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

For $K _{2} Cr _{2} O _{7}$

$\therefore \frac{N_{1}}{0.04} =\frac{294}{49}$

$N_{1} =\frac{294}{49} \times 0.04=0.24$

For $KMnO _{4}, \frac{N_{2}}{0.03} =\frac{158}{31.6}$

$N_{2} =\frac{158}{31.6} \times 0.03=0.15$

Now, from normality equation

$N_{1} V_{1}=N_{2} V_{2}$

$\Rightarrow 0.24 \times 50=0.15 \times V_{2}$

$V_{2}=\frac{0.25 \times 50}{0.15}=80\, cm ^{3}$