A compound containing only boron, nitrogen, and hydrogen was found to be 40.3% B, 5202% N, 7.5% H by mass. If 3.301g of this compound is dissolved in 50.00g of benzene, the solution produced freezes at 1.30 degree celsius. if the freezing point depression constant/cryoscopic constant is 5.12 degree celsius per metre and the freezing point of pure benzene is 5.48 degree celsius, what is the molecular weight of this compound?

Question

A compound containing only boron, nitrogen, and  hydrogen was found to be 40.3% B, 5202% N, 7.5% H by mass. If 3.301g of this compound is dissolved in 50.00g of benzene, the solution produced freezes at 1.30 degree celsius. if the freezing point depression constant/cryoscopic constant is 5.12 degree celsius per metre and the freezing point of pure benzene is 5.48 degree celsius, what is the molecular weight of this compound?

please help! thanks in advance

Jai · Answer

This is a colligative property problem. Note that when an solute is dissolved in a solvent, the freezing point of the solvent decreases. This is called freezing depression, and the change in temperature can be calculated by
(T,orig - T,new) = Kf * m
where
T,orig = original freezing point of solvent
T,new = new freezing point
Kf = freezing point constant
m = molality (mol solute / kg solvent)

In the problem,
Freezing Point of Benzene = 5.48 C
The Kf for benzene = 5.12 C / m (I think in the problem, m is not 'meter' but it's molality)
We're looking for MW of the compound. Substituting,
(T,orig - T,new) = Kf * m
5.48 - 1.3 = 5.12 * (3.301 g / MW) / 0.05 kg benzene
4.18 * 0.05 = 16.901 / MW
MW = 16.901 / (4.18 * 0.05)
MW = 80.87 g / mol

Hope this helps :3