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A 40 mL of a mixture of $H_{2}$ and $O_{2}$ at 18° C and 1 atm pressure was sparked so that the formation of water was complete. The remaining pure gas had a volume of 10 mL at 18° C and 1 atm pressure. If the remaining gas was $H_{2}$ , the mole fraction of $H_{2}$ in the 40 mL mixture is :

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0. 75

0.65

0.85

0.5

answer is A.

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Detailed Solution

In the given reaction, we have

$\begin{array}{l} H_{2} + \frac{1}{2} O_{2} ⟶ H_{2} O \\ \begin{array}{cl} x & y \\ x - 2 y & - \\ x + y = 40; & x - 2 y = 10 \end{array} \\ x = 30; x_{H_{2}} = \frac{30}{40} = 0.75 \end{array}$

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