Answer:
According to Henry's Law, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid.
[tex]\sf \dfrac{S_1}{S_2} = \dfrac{P_1}{P_2} \\ [/tex]
Given :
A gas had a solubility of 0.2 g/L at 2.5 atm
- [tex]\sf S_1 [/tex] = 0.2 g/L
- [tex]\sf P_1 [/tex] = 2.5 atm
We need to calculate the pressure of a 1.0 L sample that contains 14 g/L.
- [tex]\sf S_2 [/tex] = 14 g/L
- [tex]\sf P_2 [/tex] = ?
Solving for the unknown pressure [tex](\sf P_2)[/tex]:
[tex] \dfrac{0.2}{14} = \dfrac{2.5}{ \sf{{P_2}}}\\ \\ 0.01429 = \dfrac{2.5}{ \sf{{P_2}}} \\ \\ {\sf {P_2 }}= \dfrac{2.5}{0.01429} \\ \\ {\sf {P_2 }}= 174 \ \sf{ atm} [/tex]
Therefore, the pressure of the 1.0 L sample containing 14 g/L is 175 atm.
