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

Step-by-step 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}}=(78)/(100)=0.78`

Putting in the values we get:

`p_1=0.78* 749mmHg=584mmHg`

The partial pressure of nitrogen will be 584 mmHg