Question

The Henry’s Law constant for the solubility of $ {N_2} $ gas in water at 298 $ K $ is $ 1 \times {10^5}{\text{atm}} $ . The mole fraction of $ {N_2} $ in air is $ 0.8 $ . Calculate the number of moles of $ {N_2} $ dissolved in 10 moles of water at 298 $ K $ and $ 5{\text{ atm}} $ .
(A) $ 5 \times {10^{ - 4}} $
(B) $ 3 \times {10^{ - 5}} $
(C) $ 4 \times {10^{ - 4}} $
(D) $ 4 \times {10^{ - 5}} $

Answer 1

Hint: The Henry’s Law states that the amount of dissolved gas in a liquid is proportional to its partial pressure above the liquid. We shall use the formula given to calculate the mole fraction and thus, the moles of nitrogen.

Formula used: $ P = {K_H}X $
where $ P $ is the pressure of a gas in $ {\text{atm}} $ in a particular solvent at a fixed temperature, $ k $ is Henry's law constant, and $ X $ is the mole fraction of the gas.

Complete step by step answer:
The partial pressure of nitrogen is equal to $ {P_{{N_2}}} $ = $ {P^0} $ $ {X_{{N_2}}} $ , according to the formula of partial pressure, where $ {X_{{N_2}}} $ is the mole fraction of nitrogen in water.
Therefore, $ {P_{{N_2}}} = 5 \times 0.8 = 4 $ ,
where the total air pressure is 5 and the mole fraction of $ {N_2} $ in air is $ 0.8 $ .
 From Henry’s Law, $ P = {K_H}X $ , therefore
 $ {n_{{N_2}}} = {n_{{H_2}O}} \times \dfrac{{{P_{{N_2}}}}}{{{K_H}}} $
So, $ {n_{{N_2}}} = 10 \times \dfrac{4}{{1 \times {{10}^5}}} $ ,
where no. of moles of water = 10 moles and Henry’s Law constant for the solubility of $ {N_2} $ gas in water at 298 $ K $ is $ 1 \times {10^5} $ .
Therefore, $ {n_{{N_2}}} = 4 \times {10^{ - 4}}{\text{ atm}} $ .
So, the correct answer is option C.

Note:
The Henry’s Law constant for the solubility of a gas in a solvent is highly temperature dependent and when the temperature of the system changes the constant also changes. Solubility of gases usually decreases with the increase in temperature with some exceptions. However for aqueous solutions, the Henry’s Law constant goes through a minimum.