A solution containing 2 g of glucose (M = 180 g mol-1) in 100 g of water…

A solution containing 2 g of glucose (M = 180 g mol-1) in 100 g of water is prepared at 303 K. If the vapour pressure of pure water at 303 K is 32.8 mm Hg, what would be the vapour pressure of the solution?

Ans.

To find the vapor pressure of the solution, we can use Raoult’s law, which states that the vapor pressure of a solution is directly proportional to the mole fraction of the solvent in the solution.

The mole fraction of the solvent (water) in the solution can be calculated using the formula:

�solvent=moles of solventtotal moles in solutionXsolvent​=total moles in solutionmoles of solvent​

First, we need to find the number of moles of glucose and water in the solution.

1. Moles of glucose: $\text{Mass of glucose}=2\text{g}$ $\text{Molar mass of glucose}=180\text{g/mol}$ Number of moles of glucose=MassMolar mass=2 g180 g/mol=0.0111 molNumber of moles of glucose=Molar massMass​=180g/mol2g​=0.0111mol

2. Moles of water: $\text{Mass of water}=100\text{g}$ $\text{Molar mass of water}=18\text{g/mol}$ Number of moles of water=MassMolar mass=100 g18 g/mol=5.56 molNumber of moles of water=Molar massMass​=18g/mol100g​=5.56mol

Now, we can calculate the mole fraction of water: �solvent=5.56 mol5.56 mol+0.0111 mol≈0.9998Xsolvent​=5.56mol+0.0111mol5.56mol​≈0.9998

Now, according to Raoult’s law, the vapor pressure of the solution (�solutionPsolution​) is equal to the mole fraction of the solvent multiplied by the vapor pressure of the pure solvent (�solventPsolvent​).

�solution=�solvent×�solventPsolution​=Xsolvent​×Psolvent​ �solution=0.9998×32.8 mm HgPsolution​=0.9998×32.8mm Hg �solution≈32.78 mm HgPsolution​≈32.78mm Hg

So, the vapor pressure of the solution is approximately $32.78\text{mm Hg}$.