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 
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
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