The partial pressure of hydrogen in a flask containing $2 g$ of hydrogen and $32 g$ of sulphur dioxide is:

1 / 16^{t h}

of the total pressure

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1 / 9^{t h}

of the total pressure

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2 / 3^{r d}

of the total pressure

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1 / 8^{t h}

of the total pressure

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Solution

The correct option is C $2 / 3^{r d}$ of the total pressure
$n_{H_{2}} = \frac{2 g}{2 g/mol} = 1 mol$
and $n_{S O_{2}} = \frac{32 g}{64 g/mol} = 0.5 mol$
So, partial pressure willl be:
$∴ P_{H_{2}} = \frac{n_{H_{2}}}{(n_{H_{2}} + n_{S O_{2}})} \times P_{T} = \frac{1}{(1 + 0.5)} \times P_{T} = \frac{2}{3} P_{T}$
Therefore, the partial pressure of hydrogen is ${2 / 3}^{r d}$ of total pressure.

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