The partial pressure of ethane over a solution containing 6.56 x 10-3 g of ethane is 1 bar. If the solution contains 5.00 x 10-2 g of ethane, then what shall be the partial pressure of the gas?
According to Henry’s law
.m = k x p
Substituting the given values in the above equation.
We get
6.56 x10-3 = k x 1
Or
k = 6.56 x10-3
Now when m = 5 x 10-2,
Then again substituting the given values in Henry’s law equation, we get
5 x 10-2 = 6.56 x10-3 x p
Or
p = 7.62 bar
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Welcome to the NCERT Solutions for Class 12 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 2 , Question 13: The partial pressure of ethane over a solution containing 6.56 x 10-3 g of ethane is 1 bar. If the s....
Comments
Hey frnds this ans is correct but this is not the correct way to solve. To solve this make two equation taking no. Of moles of other component in solution as n and Henry's constant as k and partial pressure of ethane in second solution as x .While solving both equations Henry's constant and no. Of moles of other component get cancelled and u get x which is required ans.
Will you explain it in a more elaborate way?
p pressure is directly proportional to mole fraction of gas then how it becomes mass??? can anybody explain???
Partial pressure is directly proportional to the mass of the gas
P=Kh*x....where x is mole fraction then how we can replace it by mass?.... ðð
Isn't it kh*m=p. They have given it in the opposite way
I could not undrstnd that Henry\'s law states that partial pressure in vapour phase is directly proportinal to mole fraction ,then how Can it be apply directly to the given mass by replacing the mole fraction..? That is done here
Can we solve it with pie = CRT?