Using the Kf and Kb equations with electrolytes

Question

The normal freezing point of a certain liquid X is 4.00°C, but when 63.7 g of barium hydroxide(Ba(OH)2) are dissolved in 600.g of X the solution freezes at 2.8°C instead. Use this information to calculate the molal freezing point depression constant Kf of X. Be sure your answer is rounded to the correct number of significant digits.

Answer 1

We want to use rhe following equation:

ΔT = imK

ΔT = change in freezing point = 4.00 -2.8 = 1.2^o

I = vant Hoff factor = 3 for Ba(OH)2 since it ionizes into 3 particles

m = molality = moles Ba(OH)2 / kg X (see below)

K = freezing point depression constant = ?

calculation of m:

molar mass Ba(OH)2 = 171 g/mol

moles = 63.7 g x 1 mol / 171 g = 0.373 moles Ba(OH)2

kg of X = 600 g x1 kg/1000 g = 0.600 kg

m = 0.373 mol / 0.600 kg = 0.622 molal

Now, solving for K we have…

1.2^o = (3)(0.622 m)(K)

K = 0.643

Rounding to 2 sig figs (based on the final temp reading of 2.8) and introducing a minus sign since it is a depression or reduction in freezing poit, we get a value of

K = -0.64^o/m