Using the Kf and Kb equations with electrolytes

Question

A certain liquid X has a normal freezing point of 7.40°C and a freezing point depression constant Kf=2.41°C•kf•mol^-1.A solution is prepared by dissolving some ammonium sulfate((NH4)2SO4) in 700.g of X.This solution freezes at 5.4°C.Calculate the mass of (NH4)2SO4 that was dissolved. Be sure your answer is rounded to the correct number of significant digits.

Answer 1

ΔT = imK

ΔT = change in freezing point = 7.4 - 5.4 = 2.0^o

i = can’t Hoff factor = 3 since (NH4)2SO4 dissociates into 3 particles

m = molality = moles solute / kg solvent (see below)

K = freezing constant for X = 2.41^o/m

calculation of molality (m):

ΔT = imK

m = ΔT/(i)(K)

m = (2.0) / (3)(2.41)

m = 0.277 mole (NH4)2SO4 / kg X

To get grams we use molar mass of (NH4)2SO4 of 132 g/mol and molality:

mass of (NH4)2SO4 = 0.277 mol/kg x 0.700 kg = 0.194 mol x 132 g/mol = 25.6 g = 26 g (2sig figs)

Answer 2

The equation for the FPD is ΔT = -k_fpdm*i where m is the molality of the solute and iis the vant'Hoff factor. This is a number that accounts for number of particles and concentration-dependence of the colligative effect. Given no table of factors, you can assume that i=3 in this case as ammonium sulfate breaks up into 3 ions.

You can solve for the molality:

m = ΔT/(k_fpdi) = 2 °C/((2.41 °C kg/mole)(3)) = ,2766 molal

(grams of salt / molar mass of salt)) / .7 kg of X =.2766 molal From def'n of molality

solve for grams of salt