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To a $25 \mathrm{~mL} \mathrm{H}_2 \mathrm{O}_2$ solution, excess of acidified solution of $\mathrm{KI}$ was added. The iodine liberated required $20 \mathrm{~mL}$ of $0.3 \mathrm{~N} \mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3$ solution. The volume strength of $\mathrm{H}_2 \mathrm{O}_2$ solution is
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$4.08 \mathrm{~g} / \mathrm{L}$
$\begin{aligned} & \text { } 2 \mathrm{KI}+\mathrm{H}_2 \mathrm{SO}_4+\mathrm{H}_2 \mathrm{O}_2 \longrightarrow \mathrm{K}_2 \mathrm{SO}_4+2 \mathrm{H}_2 \mathrm{O}+\mathrm{I}_2 \\ & 2 \mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3+\mathrm{I}_2 \longrightarrow \mathrm{Na}_2 \mathrm{~S}_4 \mathrm{O}_6+2 \mathrm{NaI}\end{aligned}$
milli eq. of $\mathrm{H}_2 \mathrm{O}_2$ in $25 \mathrm{~mL}=20 \times 0.3=6$
milli eq. of $\mathrm{H}_2 \mathrm{O}_2$ in $1000 \mathrm{~mL}=\frac{6}{25} \times 1000=240$
Equivalent per litre $=\frac{240}{1000}$
Gram per litre of $\mathrm{H}_2 \mathrm{O}_2=\frac{240}{1000} \times 17=4.08 \mathrm{~g} / \mathrm{L}$
(Equivalent weight of $\mathrm{H}_2 \mathrm{O}_2=\frac{34}{2}=17$ ).
milli eq. of $\mathrm{H}_2 \mathrm{O}_2$ in $25 \mathrm{~mL}=20 \times 0.3=6$
milli eq. of $\mathrm{H}_2 \mathrm{O}_2$ in $1000 \mathrm{~mL}=\frac{6}{25} \times 1000=240$
Equivalent per litre $=\frac{240}{1000}$
Gram per litre of $\mathrm{H}_2 \mathrm{O}_2=\frac{240}{1000} \times 17=4.08 \mathrm{~g} / \mathrm{L}$
(Equivalent weight of $\mathrm{H}_2 \mathrm{O}_2=\frac{34}{2}=17$ ).
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