Question

The vapor pressure of pure water at 25 degrees Celsius is 23.8 torr. What is the vapor pressure of a solution prepared by dissolving 18.0 g of glucose (molecular weight = 180.0g/mol) in 95.0 g of water?

Answer 1

This question can be solved by using Raoult's law which states that the vapour pressure of a solution is equal to the product of vapor pressure of pure solvent and the mole fraction of the solvent in the solution. The vapor pressure of a solution is directly proportional to the mole fraction of solvent.

Complete answer:
The vapour pressure of the solution containing a non-volatile solute (in here glucose) can be determined by using the mole fraction of the solvent and the vapor pressure of pure solvent at a constant temperature. The formula for the same can be given as:
${p_{sol}} = {\chi _{solvent}} \times {p^0}_{solvent}$
Where, ${p_{sol}} =$ is the vapor pressure of the solution
${\chi _{solvent}} =$ mole fraction of the solvent (water)
${p^0}_{solvent} =$ vapor pressure of the pure solvent (water)
The values given to us are: ${p^0}_{water} = 23.8Torr$ . The mole fraction of solvent is not given, we’ll need to find the mole fraction of water in the solution. The formula for finding mole fraction is:
${\chi _{water}} = \dfrac{{{n_{water}}}}{{{n_{glu\cos e}} + {n_{water}}}}$
Mass of glucose = 18.0g
Molar mass of glucose = 180.0 g/mol
Mass of water = 95.0 g
Molar mass of water = 18.0 g/mol
No. of moles of Glucose present in the solution ${n_{glu\cos e}} = \dfrac{{mass}}{{molar{\text{ }}mass}} = \dfrac{{18}}{{180}} = 0.100mol$
No. of moles of water present in the solution ${n_{water}} = \dfrac{{mass}}{{molar{\text{ }}mass}} = \dfrac{{95}}{{18}} = 5.273mol$
The total no. of moles of solute and solvent ${n_{water}} + {n_{glu\cos e}} = 0.100 + 5.273 = 5.373mol$
The mole fraction of water thus will be $= \dfrac{{5.273mol}}{{5.373mol}} = 0.9814$
Substituting the values in the vapor pressure formula, we get the vapor pressure of the solution as: ${p_{solution}} = 0.9814 \times 23.8Torr = 23.4Torr$
This is the required answer.

Note:
Vapour pressure is temperature dependent. Do not confuse between vapour pressure and relative lowering of vapor pressure. The relative lowering of vapor pressure is a colligative property and can be given as: $\dfrac{{\Delta p}}{{{p^0}}} = {\chi _{solute}}$
Hence the relative lowering of vapour pressure is directly proportional to mole fraction of solute.