Answer: Partial pressure of He that would give a solubility of 0.730 M is 15.5 atm
Explanation:
Henry's law states that the amount of gas dissolved or molar solubility of gas is directly proportional to the partial pressure of the liquid.
To calculate the molar solubility, we use the equation given by Henry's law, which is:
[tex]C_{He}=K_H\times p_{liquid}[/tex]
where,
[tex]K_H[/tex] = Henry's constant =?
[tex]p_{He}[/tex] = partial pressure = 1.7 atm
Putting values in above equation, we get:
[tex]0.080=K_H\times 1.7atm\\\\K_H=0.047Matm^{-1}[/tex]
To find partial pressure of He would give a solubility of 0.730 M
[tex]0.730=0.047Matm^{-1}\times p_{liquid}[/tex]
[tex]p_{liquid}=15.5atm[/tex]
Thus partial pressure of He that would give a solubility of 0.730 M is 15.5 atm
We have that from the Question, it can be said that The partial pressure of He would give a solubility of 0.730 M is
P_2=4.7atm
From the Question we are told
At a particular temperature, the solubility of He in water is 0.080 M when the partial pressure is 1.7 atm. What partial pressure of He would give a solubility of 0.730 M
Generally the equation for constant temperature is mathematically given as
[tex]\frac{C_2}{C_1}=\frac{P_2}{P_1}\\\\Therefore\\\\P_2=\frac{P_1C_1}{C_1}\\\\P_2=\frac{0.22*1.7}{0.080}\\\\P_2=4.7atm\\\\[/tex]
Therefore
The partial pressure of He would give a solubility of 0.730 M is
P_2=4.7atm
For more information on this visit
https://brainly.com/question/19007362?referrer=searchResults
