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

Question

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 (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Milli mole of hypo =10 So percentage =(0.0355/5) × 100=0 millimole of I 2= millimoles of Cl 2=5 % available chlorine =(5 × 10-3 × 71/5) × 100 %=7.1 %

Q. $5 g$ of bleaching powder was treated with excess of acetic acid. The $C l_{2}$ was passed through excess of $K I$ Solution. The $I_{2}$ liberated required $50 m l$ of $M /5$ hypo solution. The percentage of $C l_{2}$ in sample is

Milli mole of hypo $= 10$
So percentage $= \frac{0.0355}{5} \times 100 = 0$
millimole of $I_{2} =$ millimoles of $C l_{2} = 5$
$%$ available chlorine $= \frac{5 \times 1 0 ^{- 3} \times 71}{5} \times 100% = 7.1%$

Q. 5g of bleaching powder was treated with excess of acetic acid. The Cl2​ was passed through excess of KI Solution. The I2​ liberated required 50ml of M/5 hypo solution. The percentage of Cl2​ in sample is

Milli mole of hypo =10 So percentage =50.0355​×100=0 millimole of I2​= millimoles of Cl2​=5 % available chlorine =55×10−3×71​×100%=7.1%

Q. $5 g$ of bleaching powder was treated with excess of acetic acid. The $C l_{2}$ was passed through excess of $K I$ Solution. The $I_{2}$ liberated required $50 m l$ of $M /5$ hypo solution. The percentage of $C l_{2}$ in sample is

Milli mole of hypo $= 10$
So percentage $= \frac{0.0355}{5} \times 100 = 0$
millimole of $I_{2} =$ millimoles of $C l_{2} = 5$
$%$ available chlorine $= \frac{5 \times 1 0 ^{- 3} \times 71}{5} \times 100% = 7.1%$