Homework Answers
1. moles of CaCl2 used(n) = w/m.wt = 2.2/111 = 0.0198 mole
2. mass of water being used = v*d = 500*1 = 500 g ( density of water = 1 g/ml)
3. molality of CaCl2 solution(m) = n/W
n = 0.0198 mole
w = weight of solvent = 500 g = 0.5 kg
Molality = 0.0198/0.5 = 0.0396 m
4. vanthoff factor for the CaCl2 = 3
5. DTb = i*Kb*m
Kb = ebullioscopic constant of water = 0.512 K.Kg/mol
DTb = 3*0.512*0.0396
= 0.0608
6. DTb = Boilingpoint of solution(Ts) - Boilingpoint of solvent(T0)
0.0608 = x - 100
x = Boilingpoint of solution(Ts) = 100.0608 C
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