$A5.0c{m}^3$ Solution of $H_2{O}_2$ liberates $0.508 g$ Of Iodine from an acidified KI solution. The volume strength of $H_2{O}_2$ is 'V'. Find the value of$\frac{\mathrm{V}}{1.12}$?

Moles of iodine $=\frac{0.508}{254}$

$m_1{v}_1=m_2{v}_2$

 
 $m_1\to$ molarity of  $H_2{O}_2=SoCl$
$v_1\to$  volume of  $H_2{O}_2=SoCl$
 $m_2\to$ molarity of iodine

$v_2\to 1000ml$

$m_1\times 5=\frac{0.508}{254}\times 1000$

$m_1=0.4m$

$H_2{O}_2+2I^{-}+24H^{+}\underset{ }{\overset{ }{\to }}2H_{2}O+I$

 $n$ factor of  $H_2{O}_2=2$
Normality of  $H_2{O}_2=nfactor\times m$
 $=2\times 0.4=0.8N$
Volume strength of  $H_2{O}_2=N\times 5.6$

$=0.8\times 5.6=4.48V$

$\frac{\mathrm{V}}{1.12}=\frac{4.48}{1.12}=4$