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

Step-by-step explanation:

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

`(p^o-p_s)/(p^o)=i* \chi_(solute)`

where,

`(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:

`(400-361.8)/(400)=1*\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}=(15.0g)/(0.0955mol)=157.07g/mol`

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