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  4. 5 g of bleaching powder was treated with excess of acetic acid. The Cl 2 was passed through excess of KI Solution. The I 2 liberated required 50 ml of M / 5 hypo solution. The percentage of Cl 2 in sample is

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Q. $5\, g$ of bleaching powder was treated with excess of acetic acid. The $Cl _{2}$ was passed through excess of $KI$ Solution. The $I _{2}$ liberated required $50 \,ml$ of $M / 5$ hypo solution. The percentage of $Cl _{2}$ in sample is

Redox Reactions

Solution:

Milli mole of hypo $=10$
So percentage $=\frac{0.0355}{5} \times 100=0$
millimole of $I _{2}=$ millimoles of $Cl _{2}=5$
$\%$ available chlorine $=\frac{5 \times 10^{-3} \times 71}{5} \times 100 \%=7.1 \%$