If 0.0203 g of a gas dissolves in
1.39 L of water at 1.02 atm, at what
pressure (in atm) would you be able
to dissolve 0.146 g in the same
amount of water?

A pressure of 7.33 atm is required to dissolve 0.146 g in the same  amount of water.

Explanation:

Given-

Mass of the gas, m = 0.0203 g

Volume, V = 1.39L

Pressure, P = 1.02atm

We know,

Kh = C / Pgas                

Pgas = external pressure of the gas

Kh = Henry's law constant

C = concentration of gas in moles/l

Also,

C1/C2 = / V X V/ =

= X

0.146 = 0.0203 X

= 0.146 X 1.02 / 0.0203

= 7.33 atm

Therefore, a pressure of 7.33 atm is required to dissolve 0.146 g in the same  amount of water.