Answer to Question #265339 in General Chemistry for Sunshine Jimenez

Question

The vapor pressure of water at 28°C is 28.35 torr. Compute the vapor pressure at 28°C of a solution containing 68 g of cane sugar (C₁₂H₂₂0₁₁) in 1000 g of water?

Answer 1

For solutions that contain non-volatile solutes, the vapor pressure of the solution can be determined by using the mole fraction of the solvent and the vapor pressure of the pure solvent at the same temperature.

Psol=χsolvent⋅P∘solvent , where

• Psol is the vapor pressure of the solution
• χsolvent is the mole fraction of the solvent
• P∘solvent is the vapor pressure of the pure solvent

In this case, the vapor pressure of pure water at $28^°c$ is equal to 28.35 torr. This means we determine the mole fraction of water in the solution.

Use glucose and water's respective molar masses to determine how many moles of each you have

Molar mass of cane sugar is 342.2965g/mol

$18.0g×\frac{1 mole glucose} {342.2965g} =0.0526molesglucose$

$1000g×\frac{1 mole water} {68g} =14.706moles water$

The total number of moles present in the solution will be:

$n_{total} = n_{glucose} + n_{water}$

$n_{total} =0.0526+14.706 =14.7586$

This means that the mole fraction of water will be:

$x_{water} =\frac{14.706}{14.7586} =0.9964$

Therefore, the vapor pressure of the solution will be

$P_{Sol} =0.9964 × 28.35 = 28.25 torr$