Pressure of a mixture of $4 g$ of $O_{2}$ and $2 g$ of $H_{2}$ confined in a bulb of $1 l i t r e$ at $0^{\circ} C$ is :

25.215 a t m

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31.205 a t m

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45.215 a t m

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15.210 a t m

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Solution

The correct option is C $25.215 a t m$
Pressure of a mixture of 4 g of $O_{2}$ and 2 g of $H_{2}$ confined in a bulb of 1 litre at $0^{o} C$ is 25.215 atm.
The molecular weights of $O_{2}$ and $H_{2}$ are 32 g/mol and 2 g/mol respectively.
The number of moles of $O_{2} = \frac{4 g}{32 g / m o l} = 0.125$ moles.
The number of moles of $H_{2} = \frac{2 g}{2 g / m o l} = 1$ moles.
Total number of moles $= 0.125 + 1 = 1.125$ moles
Pressure, $P = \frac{n R T}{V} = \frac{1.125 m o l \times 0.08206 L a t m / m o l K \times 273.15 K}{1 L}$

$P = 25.215 a t m$

Hence, the correct option is A.

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