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

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,

$\frac{p^o-p_s}{p^o}=i\times x_2$

where,

$\frac{p^o-p_s}{p^o}$= 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) = $\frac{\text{Given mass}}{\text {Molar mass}}=\frac{21.5g}{342g/mol}=0.0628moles$

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

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

$x_2$ = mole fraction of solute  =$\frac{0.0628}{11.1628}=5.62\times 10^{-3}$

$\frac{187.5-p_s}{187.5}=1\times 5.62\times 10^{-3}$

$p_s=186.4$

Thus the vapor pressure of the solution is 186.4 torr