A chemist prepared a 25% by wt. glucose (C6H12O6) solution in water at 29C. Determine the vapor pressure lowering of the solution. The vapor pressure of pure water at 29C is 30 Torr. ? What is the answer in this problem?

The vapor pressure of the water above the solution was decreased by #"0.968 torr"#.

To find the vapor pressure lowering, we start from Raoult's law for ideal solutions:

#P_A = chi_(A(l))P_A^"*"#

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

The change in vapor pressure above the solvent due to dissolving solute is defined as:

#DeltaP_A = P_A - P_A^"*"#

Therefore, we can derive an expression for it in terms of the solute #B#...

#color(green)(DeltaP_A) = chi_(A(l))P_A^"*" - P_A^"*"#

#= (chi_(A(l)) - 1)P_A^"*"#

#= (1 - chi_(B(l)) - 1)P_A^"*"#

#= color(green)(-chi_(B(l))P_A^"*")#

Mol fractions are always positive, so vapor pressure always decreases for dissolution of a nonvolatile solute into a solvent.

The percent by mass was given as #25%#, so for every #"100 g solution"# there is #"25 g glucose"#. From this, the mols of each substance are:

#"25 g glucose" xx "1 mol"/"180.156 g glucose" = "0.1388 mols"#

#"100 g solution" = "75 g water" + "25 g glucose"#

#=> "75 g water" xx "1 mol"/"18.015 g water" = "4.163 mols water"#

Therefore, we can calculate the mol fraction of solute:

#chi_(solute) = "0.1388 mols glucose"/("0.1388 mols glucose" + "4.163 mols water")#

#= 0.0323#

As a result,

#color(blue)(DeltaP_A) = -(0.0323)("30 torr")#

#=# #color(blue)(-"0.968 torr")#