Suppose you dissolved 0.123 gram of pentane in 2.493 grams of p-xylene and measured a freezing point depression of 2.88 degrees celcius for the solution. calculate the molar mass of pentane using this data and the value for kf that you calculated in question 1
i got .829 mol/kg for question 1

The formula for freezing point depression is:

ΔT = Kf * m         --->1

Where,

ΔT = change in temperature = 2.88 degrees Celcius

Kf = freezing point molar constant of solvent

m = molality (moles solute/mass solvent)

First we calculate for molality since we are given the mass of solute and solvent.

Molar mass of pentane = 72.15 g / mol

molality m= (0.123 g / 72.15 g / mol) / (2.493 x 10^-3 kg)

m = 0.684 molal

Going back to equation 1:

ΔT = Kf * m

2.88 = Kf * 0.684

Kf = 4.21 degC / molal

Value for Kf in question 1 given that m = 0. 829 mol/kg:

2.88 = Kf * 0.829

Kf = 3.47 degC / molal

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