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 = [tex]mass_{1}[/tex]/ V X V/ [tex]mass_{2}[/tex] = [tex]\frac{P_{1} }{P_{2} }[/tex]
[tex]mass_{2}[/tex] = [tex]mass_{1}[/tex] X [tex]\frac{P_{2} }{P_{1} }[/tex]
0.146 = 0.0203 X [tex]\frac{P_{2} }{1.02}[/tex]
[tex]P_{1}[/tex] = 0.146 X 1.02 / 0.0203
[tex]P_{2}[/tex] = 7.33 atm
Therefore, a pressure of 7.33 atm is required to dissolve 0.146 g in the same amount of water.
