Question

$5.0c{m}^{3}of{H}_2{O}_2$ liberates 0.508 g of iodine from an acidified KI solution. The strength of $H_2{O}_2$ solution in terms of volume strength at STP is:

• A. 6.48 volumes
• B. 4.48 volumes
• C. 7.68 volumes
• D. none of these

Answer 1

The correct option is B 4.48 volumes

Moles of iodine= $\frac{0.508}{254}$ moles

Now, $M_1{V}_1=M_2{V}_2$

$M_1⟶$ Molarity of $H_2{O}_2$ solution

$\therefore$ $V_1⟶$ Volume of $H_2{O}_2$

$M_2⟶$ Molarity of iodine

$V_2⟶$ $1000$ $ml$

$\therefore$ $M_1\times 5=\frac{0.508}{254}\times 1000$

$\therefore$ $M_1=0.4$ $M$

The reaction can be given as,

$H_2{O}_2+2I^{-}+2H^{+}⟶2H_{2}O+I_2$

$\therefore$ n-factor for $H_2{O}_2=2$

$\therefore$ Normality of $H_2{O}_2=2\times 0.4=0.8N$

$\therefore$ Volume strength of $H_2{O}_2=0.8\times 5.6$

=$4.48$ volumes