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?

The molar mass of unknown molecule is 157.07 g/mol

Explanation:

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

where,

= relative lowering in vapor pressure

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

= mole fraction of solute = ?

= vapor pressure of pure acetone = 400 torr

= vapor pressure of solution = 361.8 torr

Putting values in above equation, we get:

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

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

Moles of unknown molecule = 0.0955 moles

Mass of unknown molecule = 15.0 grams

Putting values in above equation, we get:

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

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