Question

To a $25ml$ $H_2{O}_2$ solution, excess of acidified solution of $KI$ was added. The iodine liberated required $20ml$ of $0.3N$ ${Na}_2{S}_2{O}_3$ solution. The volume strength of $H_2{O}_2$ solution is:

• A. $1.344g/L$
• B. $3.244g/L$
• C. $5.4g/L$
• D. $4.08g/L$

Answer 1

The correct option is D $4.08g/L$

$2KI+H_2{SO}_4+H_2{O}_2⟶K_2{SO}_4+2H_{2}O+I_2$

$2{Na}_2{S}_2{O}_3+I_2⟶{Na}_2{S}_4{O}_6+2NaI$

milli eq. of $H_2{O}_2$ in $50ml$ $=$ milli eq. of $I_2$$=$ milli eq. of ${Na}_2{S}_2{O}_3$

milli eq. of $H_2{O}_2$ in $25ml$ $=20\times 0.3=6$

milli eq. of $H_2{O}_2$ in $1000ml=\frac{6}{25}\times 1000=240$

Equivalent per litre $=\frac{240}{1000}$

Gram per litre of $H_2{O}_2$ $=\frac{240}{1000}\times 17=4.08g/L$

$($Equivalent weight of $H_2{O}_2=\frac{34}{2}=17$)

Hence, option $D$ is correct.