Question

If atmospheric pressure on a certain day is 749 mmHg, what is the partial pressure of nitrogen, given that nitrogen is about 78% of the atmosphere?165 mm Hg 760 mm Hg 600 mm Hg 584 mm Hg 749 mm Hg

Answer 1

Answer: 584 mm Hg

Explanation:

According to Raoult's law, the partial pressure of a component at a given temperature is equal to the mole fraction of that component multiplied by the total pressure.

$p_1=x_1P$

where, x = mole fraction

$P$ = total pressure = 749 mmHg

$x_{N_2}=\frac{\text {moles of }N_2}{\text {total moles}}=\frac{78}{100}=0.78$

Putting in the values we get:

$p_1=0.78\times 749mmHg=584mmHg$

The partial pressure of nitrogen will be 584 mmHg