A gas mixture containing 8.0\text{ g} of oxygen (\text{O}_2) and 7.0\text{ g} of nitrogen … — JAMB UTME

A gas mixture containing $8.0\text{ g}$ of oxygen ( $\text{O}_2$ ) and $7.0\text{ g}$ of nitrogen ( $\text{N}_2$ ) is collected over water at a total pressure of $750\text{ mmHg}$ . If the saturated vapor pressure of water at the collection temperature is $30\text{ mmHg}$ , what is the partial pressure exerted by oxygen in the dry gas mixture? (Molar masses: $\text{O}_2 = 32\text{ g/mol}$ , $\text{N}_2 = 28\text{ g/mol}$ )

$360\text{ mmHg}$ Answer
B
$375\text{ mmHg}$
C
$390\text{ mmHg}$
D
$400\text{ mmHg}$

Answer

360\text{ mmHg}

The total pressure exerted by a gas collected over water is the sum of the partial pressures of the dry gases and the saturated vapor pressure of water. Subtracting the vapor pressure of

30\text{ mmHg}

from the total pressure of

750\text{ mmHg}

yields a dry gas pressure of

720\text{ mmHg}

. Since

8.0\text{ g}

\text{O}_2

(

0.25\text{ mol}

) and

7.0\text{ g}

\text{N}_2

(

0.25\text{ mol}

) give equal mole fractions of

0.50

each, the partial pressure of oxygen is half of the dry total pressure, which equals

360\text{ mmHg}

Step-by-Step Solution

Calculate the dry gas mixture total pressure by subtracting aqueous tension from total pressure.

P_{\text{dry}} = P_{\text{total}} - P_{\text{water}} = 750\text{ mmHg} - 30\text{ mmHg} = 720\text{ mmHg}

Gas collected over water contains water vapor, so total pressure is the sum of dry gas pressure and vapor pressure (Dalton's Law).

Calculate the number of moles of each gas in the mixture.

n(\text{O}_2) = \frac{8.0}{32} = 0.25\text{ mol}

n(\text{N}_2) = \frac{7.0}{28} = 0.25\text{ mol}

Moles are required to determine the mole fraction of oxygen.

Determine the mole fraction of oxygen in the dry mixture.

\chi(\text{O}_2) = \frac{0.25\text{ mol}}{0.25\text{ mol} + 0.25\text{ mol}} = 0.50

Mole fraction represents the ratio of moles of a specific gas component to total dry moles.

Calculate the partial pressure of oxygen.

P(\text{O}_2) = \chi(\text{O}_2) \times P_{\text{dry}} = 0.50 \times 720\text{ mmHg} = 360\text{ mmHg}

By Dalton's Law of Partial Pressures, the partial pressure of a gas is equal to its mole fraction multiplied by the dry total pressure.

Key Concept

Dalton's Law of Partial Pressures and Gas Collection Over Water

Estimated Time:1m 30s

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