Question

The solubility limit of CaCl2*2H2O in boling water is 152g CaCl2*2H2O in 100g water. If you dissolved enough CaCl2*2H2O in water to reach this solubility limit, would the water boil if the water was initially 20.0 C? The answer is YES... I need the equation to prove it. How???

Answer #1

Moles of CaCl2.2H2O =

=1.03 mol

Mass of water = 100 g =0.1 kg

Molality = moles/kg = 1.03/0.1 = 10.3 m

K_{b} of water =0.512

Total number of ions CaCl2 can produce = i = 3 (Ca^{2+},
and 2Cl^{-} ions)

Formula to find elevation in boiling point

**elevation in boiling point = 100 + 15.8 = 115.8
^{o}C**

**So water boils and boiling temperature is = 115.8
^{o}C**

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