A solution of the sugar mannitol (mol wt 182.2) is prepared by adding 54.66 g of

Question

A solution of the sugar mannitol (mol wt 182.2) is prepared by adding 54.66 g of mannitol to 1000 g of water. The vapor pressure of pure liquid water is 2338 Pa at 20degree C. Mannitol is nonvolatile and does not ionize in aqueous solutions. 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 20 degree C. The observed vapor-pressure lowering of the mannitol solution above is 12.40 Pa. Calculate the activity coefficient (based on mole fraction) of water in this solution. 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.

Explanation / Answer

1. Moles of   Mannitol in 54.66 gm = mass/Molecular weight =54.66/182.2= 0.3 moles

moles of water in 1000g= 1000/18=55.55 moles

Mole fraction water= 55.55/(55.55+0.3)= 0.9946

Vapor pressure of water in solution = mole fraction water* pure component vapor presure =0.9946*2338=2325.4 Pa

lowering of vapor pressure = 2338-2325.4= 12.6 Pa

actual vapor pressure lowering = 12.4 pa

activity coefficent =   actial lowering/ lowering under ideal condition= 12.4/12.6=0.984

c) Osmotic pressure = CRT

assuming water density = 1 g/cc, mass of water= 1000/1= 1000Cc= 1L

C= 0.3/1

(0.3/1*0.0821*(20+273.15)=7.22

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