The vapour pressure of a dilute aqueous solution of Glucose is 750mm of mercury at 373K. The mole fraction of solute is?

1 Answer
Truong-Son N.
Oct 29, 2017

#chi_(j(l)) = 0.9868#

Is this the solute or the solvent? Why? Which substance is the solvent?

The "trick" to this problem is to realize that water boils at #100.0^@ "C"#, or #"373.15 K"# (is it though? You should know that by heart).

What you may not realize is that at the boiling point, the vapor pressure at the surface of the solution must match the atmospheric pressure.

Therefore, the pure vapor pressure of water at this temperature and pressure must be #"760 torr"#.

As a result, we can use it in Raoult's law for ideal solutions, which describes the reduction of vapor pressure due to addition of solute:

#P_j = chi_(j(l))P_j^"*"#

where #P_j# is the vapor pressure of solvent #j# in solution, #"*"# indicates pure solvent, and #chi_(j(l))# is the mol fraction of solvent #j# in the liquid phase.

Therefore, the mol fraction of solvent is:

#chi_(j(l)) = P_j/P_j^"*"#

#= "750 torr"/"760 torr" = 0.9868#

And thus, the mol fraction of solute, i.e. of GLUCOSE, is #color(blue)(0.0132)#. What must the mol fraction of solute AND solvent add up to?

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