RAOULT'S LAW 1- The vapor pressure of water at 25° C is 23.8 torr. What is the vapor
pressure at 25° C of a 20% aqueous solution of a non-electrolyte with a
molar mass of 121.3?

23.4 torr

Explanation:

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

=

χ

solvent

⋅

P

∘

solvent

, where

P

sol

is the vapor pressure of the solution

χ

solvent

is the mole fraction of the solvent

P

∘

solvent

is the vapor pressure of the pure solvent

In your case, you know that the vapor pressure of pure water at

25

∘

C

is equal to

23.8

torr. This means that all you have to do is determine the mole fraction of water in the solution.

As you know, mole fraction is defined as the number of moles of a component of a solution divided by the total number of moles present in that solution.

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

18.0

g

⋅

1 mole glucose

180.0

g

=

0.100 moles glucose

and

95.0

g

⋅

1 mole water

18.015

g

=

5.273 moles water

The total number of moles present in the solution will be

n

total

=

n

glucose

+

n

water

n

total

=

0.100

+

5.273

=

5.373 moles

This means that the mole fraction of water will be

χ

water

=

5.273

moles

5.373

moles

=

0.9814

Finally, the vapor pressure of the solution will be

P

sol

=

0.9814

⋅

23.8 torr

=

23.4 torr

The answer is rounded to three sig figs.