Question

Glycerol (C3H8O3, 92.1 g/mol) is a nonvolatile nonelectrolyte substance. Consider that you have an aqueous solution that contains 32.7 % glycerol by mass. If the vapor pressure of pure water is 23.8 torr at 25oC, what is the vapor pressure of the solution at 25oC? Enter your answer in units of torr to three significant figures.

Answer 1

Answer: 21.7 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 =
`\frac{\text {moles of solute}}{\text {total moles}}`

Given : 32.7 g of glycerol is present in 100 g of aqueous solution, thus (100-32.7) g = 67.3 g of water

moles of solute (glycerol) =
`\frac{\text{Given mass}}{\text {Molar mass}}=(32.7g)/(92.1g/mol)=0.355moles`

moles of solvent (water) =
`\frac{\text{Given mass}}{\text {Molar mass}}=(67.3g)/(18g/mol)=3.74moles`

Total moles = moles of solute (glycerol) + moles of solvent (water) = 0.355 + 3.74 = 4.095

`x_2` = mole fraction of solute =
`(0.355)/(4.095)=0.0867`

`(23.8-p_s)/(23.8)=1* 0.0867`

`p_s=21.7torr`

Thus the vapor pressure of the solution at
`25^0C` is 21.7 torr