16.8 litre gas containing $H_{2}$ and $O_{2}$ is formed at NTP on electrolysis of water. What should be the weight of electrolysed water

5 g

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9 g

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10 g

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12 g

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Solution

The correct option is B 9 g
$n = \frac{16.8}{22.4} = 0.75 m o l (H_{2} + O_{2})$
$2 H_{2} O \to 2 H_{2} + O_{2} 0.75 < {_{0.25}^{0.25}]}_{H_{2}}^{O_{2}}$
$2 : 1$

$2 m o l e H_{2} - 2 m o l e H_{2} O$
$0.5 m o l e H_{2} - 0.5 m o l e H_{2} O = 9 g m$

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