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Mole Fraction

In a 50 mL so...

Question

In a $50 m L$ solution of $H_{2} O_{2}$ an excess of $K I$ and dilute $H_{2} S O_{4}$ were added. The $I_{2}$ liberated required $20 m L$ of $0.1 N N a_{2} S_{2} O_{3}$ for complete reaction.
Calculate the strength of $H_{2} O_{2}$ in grams per litre.

0.25 g L^{- 1}

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0.34 g L^{- 1}

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0.68 g L^{- 1}

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0.8 g L^{- 1}

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Solution

The correct option is C $0.68 g L^{- 1}$
$2 K I (a q) + H_{2} S O_{4} (a q) + H_{2} O_{2} (a q) \to K_{2} S O_{4} (a q) + 2 H_{2} O (l) + I_{2} (g)$
$I_{2} (a q) + 2 N a_{2} S_{2} O_{3} (a q) \to N a_{2} S_{4} O_{6} (a q) + 2 N a I (a q)$

$meq. of H_{2} O_{2} in 50 ml = meq. of I_{2} = meq. of N a_{2} S_{2} O_{3}$

$meq. of H_{2} O_{2} in 50 ml = 0.1 \times 20 = 2$

$meq. of H_{2} O_{2} in 1000 ml = \frac{2}{50} \times 1000 = 40$
$Equivalents per litre = \frac{40}{1000} = 0.04$
$Since, eq. weight of H_{2} O_{2} = \frac{34}{2} = 17 ∴ Concentration of H_{2} O_{2} in grams per litre = Equivalents per litre \times eq. weight of H_{2} O_{2}$

$Concentration of H_{2} O_{2} = 0.04 \times 17 g L^{- 1} = 0.68 g L^{- 1}$

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Mole Fraction

Standard XII Chemistry

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In a 50 mL solution of H2O2 an excess of KI and dilute H2SO4 were added. The I2 liberated required 20 mL of 0.1N Na2S2O3 for complete reaction. Calculate the strength of H2O2 in grams per litre.

In a $50 m L$ solution of $H_{2} O_{2}$ an excess of $K I$ and dilute $H_{2} S O_{4}$ were added. The $I_{2}$ liberated required $20 m L$ of $0.1 N N a_{2} S_{2} O_{3}$ for complete reaction.
Calculate the strength of $H_{2} O_{2}$ in grams per litre.