Question

Two liters of acetone at 25oC dissolves 104.0 grams of acetylene, C2H2, when the partial pressure of the acetylene is 2.00 atm. (a) What is Henry’s law constant, k, for acetylene in acetone? (b) If the partial pressure of the acetylene is equal to 12.0 atm, what is the solubility of acetylene in one liter of acetone?

Answer 1

Answer : Here we have to use the Henry's law

C = HP, here C is concentration , H is henry's constant and P is partial pressure

Here , we have 104.0 gram of acetylene means number of moles

moles = given mass / molar mass = 104.0 / 26 = 4 mol

now concentration = moles / volume in L = 4 / 2 = 2M

now value of Henry's constant = C/ P = 2/2 = 1

hence the value of henry's constant = 1

b] Now again using C = HP , we get

C = 12M

Hence solubility of acetylene in one L is 12M