Question

If an ideal solution contains 3.50 mol of a non-volatile solute and 15.8 mol of solvent, and it has a vapor pressure of 22.6 torr, what is the vapor pressure of the pure solvent (in torr)

Answer 1

27.6 torr

Step-by-step explanation:

First, find the mole fraction of the solvent.

Xsolvent=15.8 mol(15.8 mol+3.50 mol)=0.81865

Next, use Rault's Law and solve for the vapor pressure of the pure solvent (P∘solvent).

PsolutionP∘solvent=XsolventP∘solvent=PsolutionXsolvent=22.6torr0.81865=27.606torr

The answer has three significant figures, so round to 27.6torr.

Answer 2

Answer: The vapor pressure of the solution is 18.5 torr

Step-by-step explanation:

As the relative lowering of vapor pressure is directly proportional to the amount of dissolved solute.

The formula for relative lowering of vapor pressure will be,

`(p^o-p_s)/(p^o)=i* x_2`

where,

`(p^o-p_s)/(p^o)`= relative lowering in vapor pressure

i = Van'T Hoff factor = 1 (for non electrolytes)

`x_2` = mole fraction of solute

Given : moles of solute = 3.50

moles of solvent = 15.8

Total moles = moles of solute + moles of solvent = 3.50 + 15.8 = 19.3

`x_2` = mole fraction of solute =
`(3.50)/(19.3)=0.181`

`(22.6-p_s)/(22.6)=1* 0.181`

`p_s=18.5torr`

Thus the vapor pressure of the solution is 18.5 torr