The normal freezing point of a certain liquidxis0.4°c, but when5.90gof ureanh22coare dissolved in450.gofx, it is found that the solution freezes at−0.5°cinstead. use this information to calculate the molal freezing point depression constantkfofx.

The molal freezing point depression constant for X is 4.12 °C/m

Explanation:

Apply the colligative property of freezing point depression:

ΔT = Kf . m . i

We assume the X lquid is non electrolytic, so i = 1

ΔT = T° freezing point of solvent pure - T° freezing point of solution

m = molality (moles of solute in 1kg of solvent)

Kf = Cryoscopic constant (the data we were asked for)

ΔT = 0.4° - (- 0.5°) = 0.9°

Let's calculate molality

Molar mass of urea = 60.06 g/m

Moles of urea = mass of urea / molar mass

5.90 g / 60.06 g/m = 0.098 moles

This moles are in 450 g of solvent, prepare a rule of three to find out the moles in 1000 g

450 g 0.098 m

1000 g ( 1000.  0.098)/450 = 0.218 m

0.9° = Kf . 0.218 m . 1

0.9° / 0.218 m = Kf → 4.12 °C/m

in the analysis of systems that involve fluid flow, we frequently encounter the combination of properties u and pv.

the correct answer to this question is 2