At 298 K and 1 atm pressure solublity of N 2 in water was found to be 6.8 × 1 0 − 4 mol / L . If partial pressure of N 2 is 0.78 atm , then the conc. of N 2 dissolved in water under atmospheric

Question

At 298 K and 1 atm pressure solublity of N 2 ​ in water was found to be 6.8 × 1 0 − 4 mol / L . If partial pressure of N 2 ​ is 0.78 atm , then the conc. of N 2 ​ dissolved in water under atmospheric conditions is Y × 1 0 − 4 M then find the value of Y .

TestHub · Accepted Answer

Concept: Henry's Law states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid.

Henry's Law: $S = k_{H} P$

Given:

$S_{1} = 6.8 \times 1 0^{- 4} m o l / L$ at $P_{1} = 1 a t m$

$P_{2} = 0.78 a t m$

First, calculate Henry's Law constant ( $k_{H}$ ):

$k_{H} = \frac{S _{1}}{P _{1}} = \frac{6.8 \times 1 0 ^{- 4} m o l / L}{1 a t m} = 6.8 \times 1 0^{- 4} m o l / (L \cdot a t m)$

Now, calculate the solubility ( $S_{2}$ ) at $P_{2} = 0.78 a t m$ :

$S_{2} = k_{H} P_{2} = (6.8 \times 1 0^{- 4} m o l / (L \cdot a t m)) \times (0.78 a t m)$

$S_{2} = 5.304 \times 1 0^{- 4} m o l / L$

The concentration is $Y \times 1 0^{- 4} M$ .

Therefore, $Y = 5.304$ .

Rounding to two decimal place, $Y = 5.3$ 0.

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