Question

A solution of the sugar mannitol (mol wt 182.2) is prepared by adding 54.66g of mannitol to 1000g of water. The vapor pressure of pure liquid water is 2338 Pa at 20⁰C. Mannitol is nonvolatile and does not ionize in aqueous solutions.

a. Assuming that aqueous mannitol solutions behave ideally, calculate the vapor pressure lowering (the difference between the vapor pressure of pure water and that of the solution) for the above solution at 20⁰C.

b. The observed vapor pressure lowering of the mannitol solution above is 12.40Pa. Calculate the activity coefficient (based on mole fraction) of water in this solution.

c. Calculate the osmotic pressure of the mannitol solution of part (b) when it is measured against pure water and compare it with the osmotic pressure of the ideal solution.

Answer 1

a) The mass of mannitol added = 54.66

Molecular weight of mannitol = 182.2

So moles of mannitol added = Mass / Mol wt = 54.66/182.2 = 0.3 moles

Moles of water = mass of water / mol wt of water = 1000 / 18 = 55.56

We know that

Vapour Pressure of solution = mole fraction of solvent X vapour pressure of pure solvent

Mole fraction of solvent = moles of water / Total moles = 55.56 / 55.56 + 0.3 = 0.99462

Pressure of solution = 0.99462 X 2338 Pa = 2325.42 Pa

So lowering = 2338 - 2325.42 = 12.58 Pa

b) If the observed lowering of vapour pressure = 12.4

So the vapour pressure of solution = 2338 - 12.4 = 2325.6

activity of water = 2325.6 / 2338 = 0.99469

Activity coeffeceint = acitivity / mole fraction of water = 0.99469 / 0.99462 = 1.000

c) The osmotic pressure = concentration X R X T

R = 0.0821

T = 200 C = 473 K

concentration = moles of mannitol / Volume of solution = 0.3 / 1000mL = 0.3 / 1 L = 0.3 M

Pressure = 0.3 X 0.0821 X 473 = 11.65 atm