Question

If .24 g of gas dissolves in 1.0 L of water at 1.5 atm of pressure, how much of the gas will dissolve if the pressure is raised at 6.0 atm? Assume the temperature is held constant

Answer 1

Henry’s Law: S=k*P ,where

S - solubility [mass of gass(g)/100 g of H2O];

k -  constant;

P - pressure;

Solubility of the unknown gas at 1.5 atm is: S= 0.24/(1000/100)=0.24*10=0.024 g/100g(H2O)

k=S/P=0.024/1.5=0.016;

Solubility at 6 atm will be:

S=k*P=0.016*6=0.096 g/100g(H2O);

So, 0.096 g of gas is in 100 g of H20, then in 1000 g of H2O the mass of the gass will be 0.96 g