College Chemistry - How to find the molar mass without the grams given

Question

Determine the molar mass of an unknown if you dissolve enough of the unknown into Benzene to make a 1.55% mass percent mixture of unknown to benzene and find that the boiling point of Benzene was raised by 2.3%

Answer 1

Colligative properties:

Normal boiling point of benzene = 80.1ºC (look it up)

∆T = imK

∆T = change in boiling point = 80.1 x 0.023 = 1.84º

i = van't Hoff factor = 1 for benzene since it is a non electrolyte

m = molality = moles solute/kg benzene = ?

K = boiling point constant for benzene = 2.53 m/ºC (look it up)

We can now solve for m

m = ∆T / (i)(K) = 1.84 / (1)(2.53)

m = 0.728 moles unknown/kg benzene

Since the solution is 1.55% by mass, we can find the mass of unknown per kg of benzene.

1.55 % by mass means 1.55 g unknown / 100 g solution. This can be expressed as follows:

1.55 g / 100 - 1.55 g benzene = 1.55 / 98.45 g benzene = 15.74 g unknown/kg benzene

m = 0.728 moles unknown/kg benzene so 15.74 g unknown = 0.728 moles

Molar mass = 15.74 g / 0.728 moles = 21.6 g/mole