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