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Salt of Strong Acid and Weak Base

5.0 cm3 of H2...

Question

$5.0 c m^{3}$ of $H_{2} O_{2}$ liberates $0.508 g$ of iodine from an acidified $K I$ solution. Calculate the strength of $H_{2} O_{2}$ solution in terms of volume strength at S.T.P.

5.72

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4.48

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3.72

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6.48

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Solution

The correct option is B $4.48$
$2 K I + H_{2} S O_{4} + H_{2} O_{2} \to K_{2} S O_{4} + I_{2} + 2 H_{2} O 34 g 254 g 5 c m^{3} 0.508 g 254 g o f I_{2} is liberated by = 34 g o f H_{2} O_{2} 0.508 g o f I_{2} is liberated by = \frac{34}{254} \times 0.508 = 0.068 g o f H_{2} O_{2}$
$5 c m^{3} o f H_{2} O_{2} contains = 0.068 g 1 c m^{3} o f H_{2} O_{2} contains = \frac{0.068}{5} = 0.0136 g$
Now $2 H_{2} O_{2} \to 2 H_{2} O + O_{2} 68 g 22400 m L a t S . T . P . 68 g o f H_{2} O_{2} give = 22400 m L o f O_{2} 1 c m^{3} o r 0.0136 g o f H_{2} O_{2} give = \frac{22400}{68} \times 0.0136 = 4.48 m L o f O_{2}$
Volume strength of $H_{2} O_{2} = 4.48$

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