Question

At 39.5 o C, the vapor pressure of pure acetone (MM = 58.08 g/mol) is 400.0 torr. If 15.0 grams of an unknown molecule is dissolved in 485.0 g acetone, the vapor pressure decreases to 361.8 torr. What is the molar mass of the solute?

Answer 1

Answer: The molar mass of unknown molecule is 157.07 g/mol

Explanation:

The equation used to calculate relative lowering of vapor pressure follows:

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

where,

$\frac{p^o-p_s}{p^o}$ = relative lowering in vapor pressure

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

$\chi_{solute}$ = mole fraction of solute = ?

$p^o$ = vapor pressure of pure acetone = 400 torr

$p_s$ = vapor pressure of solution = 361.8 torr

Putting values in above equation, we get:

$\frac{400-361.8}{400}=1\times\chi_{A}\\\\\chi_{A}=0.0955$

This means that 0.0955 moles of unknown molecule is present in the solution

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$  

Moles of unknown molecule = 0.0955 moles

Mass of unknown molecule = 15.0 grams

Putting values in above equation, we get:

$0.0955mol=\frac{15.0g}{\text{Molar mass of unknown molecule}}\\\\\text{Molar mass of unknown molecule}=\frac{15.0g}{0.0955mol}=157.07g/mol$

Hence, the molar mass of unknown molecule is 157.07 g/mol