Question

1.The freezing point of water is 0.00°C at 1 atmosphere. If 10.34 grams of calcium chloride, (111.0 g/mol), are dissolved in 157.9 grams of water.

The molality of the solution is_____ m. The freezing point of the solution is______ °C.

2.

The boiling point of
**water/****H _{2}O** is

If

The molality of the solution is______ m.

The boiling point of the solution is ________ °C.

Answer #1

1) molality = (W /MW) x (1000 / W of solvent in g)

molality = (10.34 / 111) (1000 /157.9 )

molality = 0.093 x 6.33

**molality = 0.59 m**

CaCl2 ----------> Ca^{+2} + 2Cl^{-}

i = 3

Tf = i x Kf x m

Tf = 3 x 1.86 x 0.59

Tf = 3.29

freezing point of solution = 0 - 3.29 = -3.29^{0}C

**freezing point of solution
= -3.29 ^{0}C**

2) molality = (13.32 / 166) (1000 / 186.9)

molality = 0.08 x 5.3

**molality = 0.424**
m

KI ------------>
K^{+} + I^{-}

i = 2

Tb = i x Kb x m

Tb = 2 x 0.512 x 0.424

Tb = 0.43

boiling point of solution = 100 + 0.43

**boiling point of solution =
100.43 ^{0}C**

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