Chemistry, 26.07.2019 21:30 bstine6678

Dry air is 78.08% nitrogen, 20.95% oxygen and 0.93% argon with the 0.04% being other gases. if the atmospheric pressure is 760.0 torr, what is the partial pressure of nitrogen in dry air?

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Answer from: hardwick744

Explanation:

Let us assume that volume of air present is 100.

So, when 78.08% nitrogen is present then it means that volume of nitrogen is 78.08.

Similarly, volume of oxygen is 20.95, volume of argon is 0.93 and volume of other gases is 0.04.

Hence, calculate the mole fraction of given gases as follows.

Mole fraction of $N_{2}$ = $\frac{78.08}{100}$

= 0.7808

Mole fraction of $O_{2}$ = $\frac{20.95}{100}$

= 0.2095

Mole fraction of Ar = $\frac{0.93}{100}$

= 0.0093

Mole fraction of other gases = $\frac{0.04}{100}$

= 0.0004

As the mole fraction of nitrogen is 0.7808 and total pressure is 760.0 torr. Therefore, calculate the partial pressure of nitrogen as follows.

$p_{i} = x_{i} \times P_{total}$

= $0.7808 \times 760 torr$

= 593.408 torr

Therefore, we can conclude that the partial pressure of nitrogen in dry air is 593.408 torr.

Answer from: Quest

it is b that is the answer.

Answer from: Quest

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Dry air is 78.08% nitrogen, 20.95% oxygen and 0.93% argon with the 0.04% being other gases. if the a...

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