Twenty grams of a solute is added to 100g of water at 25 degrees C. The vapor pressure of pure water is 23.76mm; the vapor pressure of the solution is 22.41mm.
A.) Calculate the molecular weight of the solute.
B.) What weight of this solute is required in 100g of water to reduce the vapor pressure to half the value for pure water?
Answers
Vapour pressure of solution = Mole fraction of solvent × vapour pressure of pure solvent.
Mole fraction of solvent = Moles of solvent / Total moles of solvent and solute
Question A:
Let M be the molar mass of solute.
Mole ratio of solvent :
(100/18) / (100/8) + (20/M)
Substituting in the formula :
100 / 18 = 5.555
{5.555 / (5.555 + 20 / M)} × 23.76 = 22.41
131.99 = 22.41 [5.555 + 20 / M]
131.99 = 124.49 + 448.20/ M
131.99 M = 124.49 M + 448.20
131.99 M - 124.49 M = 448.20
7.5 M = 448.20
M = 59.76 grams / mol
Question B:
Given that the vapour pressure of the solution reduces to half the value of vapour pressure of pure water we have %
23.76/2 = 11.88
Let X be the mass of solute then
11.88 = [5.555 / 5.555 + X/59.76] × 23.76
65.99 + 11.88X / 59.76 = 131.99
3943.56 + 11.88X = 7887.72
11.88 X = 7887.72 - 3943.56
11.88 X = 3944.16
X = 3944.16 / 11.88
X = 332 g
The required weight is : 332 - 20 = 312 g
Answer is 312 g.
Formula :
Vapour pressure of solution = Mole fraction of solvent × vapour pressure of pure solvent.
Mole fraction of solvent = Moles of solvent / Total moles of solvent and solute
Question A:
Let M be the molar mass of solute.
Mole ratio of solvent :
(100/18) / (100/8) + (20/M)
Substituting in the formula :
100 / 18 = 5.555
{5.555 / (5.555 + 20 / M)} × 23.76 = 22.41
131.99 = 22.41 [5.555 + 20 / M]
131.99 = 124.49 + 448.20/ M
131.99 M = 124.49 M + 448.20
131.99 M - 124.49 M = 448.20
7.5 M = 448.20
M = 59.76 grams / mol
Question B:
Given that the vapour pressure of the solution reduces to half the value of vapour pressure of pure water we have %
23.76/2 = 11.88
Let X be the mass of solute then
11.88 = [5.555 / 5.555 + X/59.76] × 23.76
65.99 + 11.88X / 59.76 = 131.99
3943.56 + 11.88X = 7887.72
11.88 X = 7887.72 - 3943.56
11.88 X = 3944.16
X = 3944.16 / 11.88
X = 332 g
The required weight is : 332 - 20 = 312 g
Answer is 312 g.