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A gaseous mix...

Question

A gaseous mixture was prepared by taking equal moles of $N_{2}$ and $O_{2}$ at room temperature. If the total pressure of the gaseous mixture was found to be 1 atm, then the partial pressure of nitrogen $(N_{2})$ in the gaseous mixture is:

A

0.6 atm

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B

1 atm

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C

0.5 atm

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D

0.9 atm

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Solution

The correct option is C 0.5 atm
Given
$∵ n_{O_{2}} = n_{N_{2}}$
P = Total pressure = 1 atm
$P_{N_{2}} = \frac{n_{N_{2}}}{n_{O_{2}} + n_{N_{2}}} \times 1 = \frac{n_{N_{2}}}{2 n_{N_{2}}} \times 1 = \frac{1}{2} = 0.5 a t m$

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