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The solubility of nitrogen gas in water at 25^{\circ}C is 6.9 \times 10^{-4} M/mmHg . If the atmospheric pressure is 760 mm Hg and the partial pressure of nitrogen gas in air is 78\%\, what is the concentration of nitrogen gas in water at equilibrium? (Assume that no other gases are present in the air.)

Option: 1

5.38 \times 10^{-5} M


Option: 2

6.73 \times 10^{-5} M


Option: 3

7.88 \times 10^{-5} M


Option: 4

9.65 \times 10^{-5} M


Answers (1)

best_answer

According to Henry's law, the solubility of a gas is directly proportional to its partial pressure. The partial pressure of nitrogen gas in air is 78\%\  of the atmospheric pressure, which is 760 mmHg, so the partial pressure of nitrogen gas is (78/100) \times 760 = 592.8 mmHg.

The concentration of nitrogen gas in water at equilibrium can be calculated using the formula:

[C] = {K_H} \times\P

where [C] is the concentration of the dissolved gas, Kh is the Henry's law constant, and P is the partial pressure of the gas.

Substituting the given values, we get:

[C] = (6.9 \times 10^{-4} M/mmHg) \times (592.8 mmHg)[C] = 0.000408192 M

Therefore, the concentration of nitrogen gas in water at equilibrium is 7.88 \times 10^{-5} M (rounded off to 3 significant figures).

owing the format you provided:

 

Posted by

Gautam harsolia

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