Question

What is the volume (in $mL$ ) of $20 \,vol\, H _{2} O _{2}$ required to completely react with $500\, mL$ of $0.02\, M$ acidified $KMnO _{4}$ solution?

• A. 14.0
• B. 7.0
• C. 28.0
• D. 42.0

Answer 1

Answer: (A)

Given,

Volume of acidified $KMnO _{4}$ solution $=500 \,mL$

Molarity of acidified $KMnO _{4}$ solution $=0.02\, M$

Volume strength of $H _{2} O _{2}=20\, vol$

$\because$ Normality $=\frac{\text { Volume strength }}{\text { Equivalent weight }}$

$\therefore $ Normality for $H _{2} O _{2}=\frac{20}{5.6}=3.57\, N$

$\because$ For $KMnO _{4}$, reaction is in acidic medium, thus valence factor is $5$.

$Mn ^{7+}+5 e^{-} \longrightarrow Mn ^{2+}$

Thus, normality for $KMnO _{4}=$ Molarity $\times 5$

Now, applying normality equation,

$N_{1} V_{1}\left( H _{2} O _{2}\right) =N_{2} V_{2}\left( KMnO _{4}\right) $

$3.57 \times V_{1} =0.02 \times 5 \times 500$

$V_{1} =\frac{50}{3.57}=14.0\, mL$