queston in description

Question

A solution contains 120.5g of CaCI₂ dissolved in 517 grams of water.

 

A. if water boils at 97.5 at what temperature will this solution boil?

 

B.If water freezes at 0.7 C at what temp. with this solution freeze?

 

 

I need help solving 

 

Answer 1

Colligative Properties:

∆T = imK

∆T = ?

i = 3 for CaCl2

m = 120.5 g/111g /0.517 kg = 2.0998 m

K = 0.512ºC/m

Solving for ∆T:  ∆T = (3)(2.0998)(0.512) = 3.23º 

A:  The solution will now boil at 97.5 + 2.32 = 100.7 degrees C

 

For Freezing, ∆T = (3)(2.0998)(-1.86) = 11.7ºC

B:  The solution will now freeze at 0.7 - 11.7 = 11.0 degrees C