Question

$10mL$ of $H_2{O}_2$ weighs $10.205g$. The solution was diluted to $250mL$, $25mL$ of which required $35.8mL$ of a decinormal solution of $KMn{O}_4$. Calculate the weight in grams of $H_2{O}_2$ in $100mL$ and also the volume strength of the solution.

Answer 1

$10ml$ of solution containing $10.25g$ of $H_2{O}_2$ diluted to $250ml$ which will contain same mass of $H_2{O}_2$

Now $25ml$ is extracted which will contain $=\frac{10.25}{250}\times 25=1.025g$

$H_2{O}_2+KMn{O}_4⟶O_2+{Mn}^{2+}$

$1.025g$ $35.8ml$

$(v.f=2)$ $\frac{1}{10}N$

meq of $H_2{O}_2$ in $25ml=$ meq of $KMn{O}_4$ used

$\frac{wt}{34/2}\times 1000=35.8\times \frac{1}{10}=3.58$

wt of $H_2{O}_2$ in $25ml=\frac{3.58\times 17}{1000}=0.06086g$

wt of $250ml$ (diluted)$=0.06086\times 10=0.6086g$

wt of original $10ml$ solution for $H_2{O}_2=0.6086g$

wt in original $100ml$ solution for $H_2{O}_2=0.6086\times 10g=6.086g$

Volume strength $=$ normality of solution $\times 5.6$

$=\frac{6.086/17}{100}\times 1000\times 5.6=20$ volume

($\therefore$ $5.6N$ $H_2{O}_2$ solution $=1$ volume of $H_2{O}_2$ solution)