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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

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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

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