A $1.0 g$ sample of air consists of approximately $0.76 g$ of nitrogen and $0.24 g$ of oxygen. This sample occupies a $1.0 L$ vessel at $20^{o} C$ . Then:

The partial pressure of

N_{2}

0.65 a t m

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The partial pressure of

O_{2}

0.36 a t m

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The total pressure is

0.83 a t m

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The total pressure is

1.05 a t m

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Solution

The correct option is B The total pressure is $0.83 a t m$
conclusion: hence the option (C) is correct.
The data needed for use are
$P_{N_{2} =} ? n_{N_{2}} = ? V = 1.00 L T = 293 K (20^{0} C)$

$P_{O_{2}} = ? n_{O_{2}} = ? V = 1.00 L T = 293 K (20^{0} C)$

The number of $n_{N_{2}}$ and $n_{O_{2}}$ can be obtained from the masses and molar masses (28.02g/mol for $N_{2}$ 32.00g/mol of $O_{2}$ 0

Moles of $N_{2}$ = 0.76G $N_{2}$ x = $\frac{1 m o l N_{2}}{28.02 g N_{2}} = 0.0271 m o l N_{2}$

Moles of $O_{2}$ =0.24g $O_{2}$ x= $\frac{1 m o l O_{2}}{32.00 {g O}_{2}} = 0.00750 m o l O_{2}$
using $P_{A} = \frac{n_{A} R T}{V}$
$P_{N_{2} =} 0.0271 m o l N_{2} \times \frac{0.08206 L a t m}{1 K m o l} \times \frac{293}{1.00 L} = 0.65 a t m N_{2}$

$P_{O_{2}} = 0.00750 m o l O_{2} \times \frac{0.08206 L a t m}{1 K m o l} \times \frac{293}{1.00 L} = 0.18 a t m O_{2}$

The total pressure is the sum of these partial pressure

$P = 0.65 a t m + 0.18 a t m = 0.83 a t m$

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Similar questions

Q. A sample of air contains nitrogen, oxygen and saturated with water vapour under a total pressure of

640

mm. If the vapour pressure of water at that temperature is

40

mm and mole ratio of

N_{2}

O_{2}

3 : 1

, the partial pressure of nitrogen in the sample is:

A sample of air contains nitrogen, oxygen and saturated with water vapour under a total pressure of 640 mm. If the vapour pressure of water at that temperature is 40 mm and the molecular ratio of $N_{2} : O_{2}$ is 3 : 1. Find the partial pressure of nitrogen in the sample.