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

Hint: 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.