What is the vapor pressure of the pure solvent if the vapor pressure of a solution of 10.0 g of sucrose, `C_6H_12O_6`, in 100.0 g of ethanol is 55.0 mmHg?

The vapor pressure of the pure solvent will be equal to 55.7 mmHg.

You're dealing with a solution of sucrose, which is your solute, dissolved in ethanol, which is your solvent.

Since sucrose is a non-volatile solute, the vapor pressure of the solution can be expressed using the mole fraction of the solvent - this is known as Raoult's Law.

`P_"solution" = chi_"solvent" * P_"solvent"^@`, where

`P_"solution"` - the vapor pressure of the solution;
`chi_"solvent"` - the mole fraction of the solvent;
`P_"solvent"^@` - the vapor pressure of the pure solvent.

The mole fraction of the solvent is defined as the number of moles of ethanol divided by the total number of moles present in the solution.

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

`10.0cancel("g") * "1 mole sucrose"/(342.3cancel("g")) = "0.0292 moles sucrose"`

and

`100.0cancel("g") * "1 mole ethanol"/(46.07cancel("g")) = "2.171 moles ethanol"`

The total number of moles will be

`n_"total" = 0.0292 + 2.171 = "2.20 moles"`

The vapor pressure of the pure solvent will thus be

`P_"solvent"^@ = P_"solution"/chi_"solvent"`

`P_"solvent"^@ = "55.0 mmHg"/((2.171cancel("moles"))/(2.20cancel("moles"))) = color(green)("55.7 mmHg")`