Question

A certain liquid X has a normal freezing point of 5.50 °C and a freezing point depression constant K =4.39 °C-kgmol. A solution is prepared by dissolving some urea (CH4N20) in 800. g of X. This solution freezes at 3.6 °C. Calculate the mass of CH4N20 that was dissolved. Be sure your answer is rounded to the correct number of significiant digits. x 6 ?

Answer 1

Δ Tf = 5.50 oC - 3.6 oC = 1.9 oC

use:
Δ Tf = Kf*mb
1.9 = 4.39 *mb
mb= 0.432802 molal


m(solvent)= 800 g
= 0.8 Kg

number of mol,
n = Molality * mass of solvent in Kg
= (0.432802 mol/Kg)*(0.8 Kg)
= 0.3462 mol

Molar mass of CH4N2O,
MM = 1*MM(C) + 4*MM(H) + 2*MM(N) + 1*MM(O)
= 1*12.01 + 4*1.008 + 2*14.01 + 1*16.0
= 60.062 g/mol

mass of CH4N2O,
m = number of mol * molar mass
= 0.3462 mol * 60.062 g/mol
= 20.8 g

Answer: 20.8 g