Question

Determine the freezing point of a solution that contains 3.05 g of copper(II) chloride (134.45 g/mol) dissolved in 50.0 grams of water.      (∆T_f = i K m) where K = 1.86 °C/m. Show your work.

	a.	0.00° C
	b.	- 2.53° C
	c.	- 1.69° C
	d.	2.53° C

Answer 1

CuCl2 dissociates into 1 Cu2+ and 2 Cl-

So, i=3 for CuCl2

Lets calculate molality first

Molar mass of CuCl2 = 1*MM(Cu) + 2*MM(Cl)

= 1*63.55 + 2*35.45

= 134.45 g/mol

mass of CuCl2 = 3.05 g

we have below equation to be used:

number of mol of CuCl2,

n = mass of CuCl2/molar mass of CuCl2

=(3.05 g)/(134.45 g/mol)

= 2.269*10^-2 mol

mass of solvent = 50.0 g

= 5*10^-2 kg [using conversion 1 Kg = 1000 g]

we have below equation to be used:

Molality,

m = number of mol / mass of solvent in Kg

=(2.269*10^-2 mol)/(0.05 Kg)

= 0.4537 molal

lets now calculate deltaTf

deltaTf = i*Kf*m

= 3.0*1.86*0.4537

= 2.5316 oC

This is decrease in freezing point

freezing point of pure liquid = 0.0 oC

So, new freezing point = 0 - 2.5316

= -2.53 oC

Answer: b