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Bleaching Powder

1.25 g of ble...

Question

$1.25 g$ of bleaching powder when treated with acetic acid and excess of $K I$ , liberated iodine which required $37.5 m L$ of $0.2 N$ sodium thiosulphate solution. Calculate the percentage of available chlorine in the sample of bleaching powder.

30.0 %

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21.3 %

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12.5 %

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9.7 %

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Solution

The correct option is B $21.3 %$
$C a O C l_{2} (a q) + 2 C H_{3} C O O H (a q) \to C a (C H_{3} C O O)_{2} + C l_{2} (g) + H_{2} O (l)$
$C l_{2} (g) + 2 K I (a q) \to 2 K C l (s) + I_{2} (g)$
$I_{2} (a q) + 2 N a_{2} S_{2} O_{3} (a q) \to N a_{2} S_{4} O_{6} (a q) + 2 N a I (a q)$

$Equivalents of N a_{2} S_{2} O_{3} = Equivalents of I_{2} = Equivalents of C l_{2}$
$∴ Equivalents of C l_{2} = \frac{0.2 \times 37.5}{1000} = 0.0075$
$Amount of C l_{2} = \frac{0.0075 \times 71}{2} = 0.266 g$
$%$ of $C l_{2} = \frac{0.266}{1.25} \times 100 % = 21.3 %$

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C a (O C l)_{2} + 2 C H_{3} C O O H \to C a (C H_{3} C O O)_{2} + H_{2} O + C l_{2}

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