The Henry’s law constant for the solubility of N 2 gas in water at 298 K is 1 × 10 - 5 a t m . The mole fraction of N 2 in air is 0. 8 . If the number of moles of N 2 of air dissolved in 10 m o l e s of water at 298 K and 5 a t m is x · 10

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The Henry’s law constant for the solubility of N 2 gas in water at 298 K is 1 × 10 - 5 a t m . The mole fraction of N 2 in air is 0. 8 . If the number of moles of N 2 of air dissolved in 10 m o l e s of water at 298 K and 5 a t m is x · 10 - 4 . Find the value of x.

Quantrex Academy · Accepted Answer

Correct answer: 4

4.00 X N 2 = P N 2 1 K H = 5 × 0 · 8 1 × 10 - 5 P N 2 1 = P a i r mole fraction of N 2 = 4 × 10 - 5 &#x2234; n N 2 n N 2 + n H 2 O = 4 &#x00D7; 10 &#x2013; 5 = n N 2 n N 2 + 10 &#x2243; n N 2 10 ∴ n = 4 × 10 – 4 Hence x = 4

The Henry’s law constant for the solubility of N 2 gas in water at 298 K is 1 × 10 - 5 a t m . The mole fraction of N 2 in air is 0. 8 . If the number of moles of N 2 of air dissolved in 10 m o l e s of water at 298 K and 5 a t m is x · 10 - 4 . Find the value of x.

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