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Standard XIII

Chemistry

Ideal Gas Equation

A 821 mL N2 g...

Question

A $821 mL N_{2} (g)$ was collected over liquid water at $300 K$ and 1 atm. If the vapour pressure of $H_{2} O$ is 30 torr then number of moles of $N_{2} (g)$ in the moist gas mixture is:

0.39

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0.032

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0.96

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0.0013

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Solution

The correct option is B $0.032$
Given : Volume, V = 821 mL = 0.821 L
Temperature, T = 300 K
Total pressure, $P_{t o t a l} = 1 atm = 760 torr$
Vapour pressure of $H_{2} O$ , $P_{H_{2} O} = 30 torr$
$P_{t o t a l} = P_{N_{2}} + P_{H_{2} O}$
$760 torr = P_{N_{2}} + 30 torr$
$P_{N_{2}} = 730 torr = \frac{730}{760} atm$
Using the ideal gas equation, moles of $N_{2} (g)$ in moist gas mixture is:
$n_{N_{2}} = \frac{P_{N_{2}} V}{R T}$

$n_{N_{2}} = \frac{(\frac{730}{760}) \times 0.821}{0.0821 \times 300} = 0.032$
Hence, the correct answer is option (b).

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Ideal Gas Equation

Standard XIII Chemistry

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A 821 mL N2 (g) was collected over liquid water at 300 K and 1 atm. If the vapour pressure of H2O is 30 torr then number of moles of N2 (g) in the moist gas mixture is:

A $821 mL N_{2} (g)$ was collected over liquid water at $300 K$ and 1 atm. If the vapour pressure of $H_{2} O$ is 30 torr then number of moles of $N_{2} (g)$ in the moist gas mixture is: