Question

derive an expression for depression of freezing point and mole fraction of the solute dissolved​

Answer 1

Answer:

2.00 g of some unknown compound reduces the freezing point of 75.00 g of benzene from 5.53 to 4.90 ∘C

. What is the molar mass of the compound?

Solution

First we must compute the molality of the benzene solution, which will allow us to find the number of moles of solute dissolved.

m=ΔTf−Kf=(4.90−5.53)∘C−5.12∘C/m=0.123m

Amount Solute=0.07500kgbenzene×0.123m1kgbenzene=0.00923msolute

We can now find the molecular weight of the unknown compound:

Molecular Weight=2.00gunknown0.00923mol=216.80g/mol

The freezing point depression is especially vital to aquatic life. Since saltwater will freeze at colder temperatures, organisms can survive in these bodies of water