Question

What is the volume of $0.5N{KMnO}_4$ required to completely react with $0.2$ mole of ${FeC}_2{O}_4$ in an acidic medium?

• A. $6l$
• B. $1.2l$
• C. $2l$
• D. $12l$

Answer 1

The correct option is B $1.2l$

Solution:- (B) 1.2 L

As we know that,

No. of gram equivalent $=n\times f=N\times V$

whereas,

$n=$ no. of moles

$f=$ valence factor

$N=$ Normality

$V=$ Volume

In acidic medium, for ferrous oxalate,

${Fe}^{+2}\to {Fe}^{+3}+e^{-}$

${C_2{O}_4}^{-2}\to +2C{O}_2+2e^{-}$

$\because$ 3 electrons are lost per molecule of ferrous oxalate, valence factor for $Fe{C}_2{O}_4$ is 3.

Given that:-

$n_{Fe{C}_2{O}_4}=0.2\text{moles}$

$N_{KMn{O}_4}=0.5N$

$V_{KMn{O}_4}=?$

$\therefore {\left(\text{gm. eq}\right)}_{Fe{C}_2{O}_4}={\left(\text{gm. eq.}\right)}_{KMn{O}_4}$

${(n\times f)}_{Fe{C}_2{O}_4}={(N\times V)}_{KMn{O}_4}$

$\Rightarrow 0.2\times 3=0.5\times V$

$\Rightarrow V=\frac{0.6}{0.5}=1.2L$

Hence, 1.2 L volume of 0.5 N $KMn{O}_4$ required to completely react with 0.2 mole of $Fe{C}_2{O}_4$ in acidic medium.