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/H₂O is 100.00 °C at 1 atmosphere.

If 13.32 grams of potassium iodide, (166.0 g/mol), are dissolved in 186.9 grams of water ...

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 = -3.29⁰C

freezing point of solution =  -3.29⁰C

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 = 100.43⁰C