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

Hi, we have to first calculate the mole fractions.

no. of moles of water = 1000g/18.0 = 55.56 mol

no. of moles of mannitol = 54.66/182.3 = 0.30 mol

mole fraction of water here is:

solution vapor pressure is:

Vapor pressure lowering is 2338 - 2326.3 = 11.7 Pa

the activity coefficient is:

Calculated vapor pressure is 2326.3 Pa while the actual is 2338-12.40 = 2325.6 Pa

so

for osmotic pressure

we need the molarity of mannitol

since the observed vapor pressure is 2325.6 Pa, we can calculate the mole fraction of water as:

we can calculate the moles of mannitol as

, this is the formula for mole fraction

so, moles of mannitol are found to be 0.279

since there is 1000 of water, it is essentially 1000 mL since the density of water is nearly 1. We can say that the molarity of mannitol solution is 0.279 M

and

= 6.71 atm

now come to the original mass we added

we added 0.30 mol of mannitol in 1000 mL water so

so you can see that the theoretical osmotic pressure is higher while the actual osmotic pressure is low. This is because the actual vapor pressure lowering is lower than the observed vapor pressure lowering.