Question

Calculate the vapor pressure of water above a solution prepared by adding 21.5 g of lactose (C12H22O11) to 200.0 g of water at 338 K. (Vapor-pressure of water at 338 K 187.5 torr.)

Answer 1

Answer: The vapor pressure of the solution is 186.4 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 : 21.5 g of lactose is present in 200.0 g of water.

moles of solute (lactose) =
`\frac{\text{Given mass}}{\text {Molar mass}}=(21.5g)/(342g/mol)=0.0628moles`

moles of solvent (water) =
`\frac{\text{Given mass}}{\text {Molar mass}}=(200.0g)/(18g/mol)=11.1moles`

Total moles = moles of solute (lactose) + moles of solvent (water) = 0.0628+ 11.1 = 11.1628

`x_2` = mole fraction of solute =
`(0.0628)/(11.1628)=5.62* 10^(-3)`

`(187.5-p_s)/(187.5)=1* 5.62* 10^(-3)`

`p_s=186.4`

Thus the vapor pressure of the solution is 186.4 torr