Question

The solubility of a gas in water at 300 K under
a pressure of 100 atmospheres is 4 x 10 kg
L-!. Therefore, the mass of the gas in kg
dissolved in 250 mL of water under a pressure
of 250 atm and 300 K is :-​

Answer 1

according to Henry's law, solubility of gas is directly proportional to pressure of gas at a constant temperature.

i.e., $S\propto P$

or, $\frac{S_1}{S_2}=\frac{P_1}{P_2}$

case 1 : The solubility of a gas in water at 300K under pressure of 100atm is 4 × 10^-3 Kg/L.

case 2 : now pressure is increased to 250 atm at same temperature i.e., 300K

now, 4 × 10^-3/S = 100/250

or, 4 × 10^-3 × 2.5 = S

or, S = 10^-2 Kg/L

hence, solution becomes 10^-2 Kg/L when pressure is increased to 250atm.

now, mass of gas in 250ml solution = solubility × volume of solution

= 10^-2 kg/L × 250ml

= 10^-2 × 10³ g/1000ml × 250ml

= 1/4 × 10 g

= 2.25 g

hence, mass of gas is 2.5g