Answer to Question #332339 in General Chemistry for harvey

Question

The vapor pressure of water at 25 ºC is 23.8 mmHg. What is the vapor pressure of a solution containing 5.50 g of nonelectrolyte sucrose (MWsucrose= 342 g/mol) in 12.8 g of water (MWwater= 18.0 g /mol) at 25.0 °C

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.   P_sol = X_solventâ‹…P⁰_{solvent​,} where     P_sol is the vapor pressure of the solution     X_solvent is the mole fraction of the solvent     P⁰_solvent is the vapor pressure of the pure solvent
Moles of sucrose = 5.50g / 342.0g/mol = 0.016 moles
Moles of water = 12.8g / 18g/mol = 0.71 moles
The total number of moles present in the solution will be:   n_total = n_sucrose +n_water   n_total ​= 0.016 + 0.71= 0.726 moles
This means that the mole fraction of water will be:   X_water = 0.71 moles / 0.726 moles = 0.978
Finally, the vapor pressure of the solution will be:   P_sol = 0.978 ⋅23.8 mmHg = 23.2784 mmHg
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