Question

$5.7$ g of bleaching powder was suspended in $500$ mL of water. $25$ mL of this suspension on treatment with $KI$ and $HCl$ liberated iodine which reacted with $24.35$ mL of $\frac{N}{10}$ $N{a}_2{S}_2{O}_3$. Calculate the percentage of available chlorine in bleaching powder.

• A. $30.33\%$
• B. $13.33\%$
• C. $66.66\%$
• D. $33.33\%$

Answer 1

The correct option is A $30.33\%$

The complete reaction is given below:

Bleaching powder$\stackrel{KI+HCl}{\to }{I}_2\stackrel{N{a}_2{S}_2{O}_3}{\to }{I}^{-}+N{a}_2{S}_4{O}_6$
The redox changes are as follows:
$I_2+2e⟶2I^{-}$
$2\left(S^{2+}\right)⟶(S^{5/2+}{)}_4+2e$
Meq. of bleaching powder $=$ Meq. of available $C{l}_2=$ Meq. of liberated $I_2=$ Meq. of $N{a}_2{S}_2{O}_3$ used
Meq. of available $C{l}_2$ in $25$ mL bleaching powder solution$=$ Meq. of $N{a}_2{S}_2{O}_4$ used $=24.35\times \frac{1}{10}$
$\therefore$ Meq. of available $C{l}_2$ in $500$ mL bleaching powder solution $=24.35\times \frac{1}{10}\times \frac{500}{25}=48.7$
or, $\frac{w}{\left(\frac{71}{2}\right)}\times 1000=48.7⟹w_{C{l}_2}=1.729g$
Percentage of available $C{l}_2$ in bleaching powder $=\left(\frac{1.729}{5.7}\right)\times 100=30.33\%$