vapor pressure curve for water

V. P

. of t

he s

olut

ion

Mole Fraction of Solvent0 1

V.P. of pure solvent

V.P.(sol’n) = (solv) V.P.(solv)

P

TNormal F.P. Normal B.P.

V.P. solid = V.P. liquid

New, Lower F.P.

T = Kfm

1.0 Atm

V.P. liquid

What is the freezing point of an aqueous solution of MgSO4 if 0.602 grams of magnesium sulfate was dissolved in 100. mL of water? (Kf for water is 1.86 K kg/mol)

T=[1.86K kg/mol][(0.602g/120.4 g/mol)/0.100kg)]

T=0.0930 K therefore Tf= -0.0930°C

m = 0.050 x 2

T = Kfmi van’t Hoff Factor

2

T=0.186 K therefore Tf= -0.186°C

….BUT WAIT!!

P

TNew, higher B.P.Normal B.P.

V.P. solid = V.P. liquid

T = Kbmi

1.0 Atm

V.P. liquid

What is the boiling point of an aqueous solution of MgSO4 if 0.602 grams of magnesium sulfate was dissolved in 100. mL of water? (Kb for water is 0.51 K kg/mol)

T= [0.51K kg/mol] [(0.0500m] 2

T=0.051 K therefore Tb= 100.051°C

T(meas.)=0.033 K therefore Tb (meas.)= 100.033°C

Therefore, the i value must only be 1.3; how come?

….BUT WAIT!!

What is the freezing point of an aqueous solution of MgSO4 if 0.602 grams of magnesium sulfate was dissolved in 100. mL of water? (Kf for water is 1.86 K kg/mol)

T=[1.86K kg/mol][(0.602g/120.4 g/mol)/0.100kg)]

T=0.0930 K therefore Tf= -0.0930°C….BUT WAIT!!

m = 0.050 x 2

T = Kfmi van’t Hoff Factor

2

T=0.186 K therefore Tf= -0.186°C

x 1.3

1.3

0.121 K -0.121° C

•Have you ever heard of ethylene glycol?

•Calculate the boiling and freezing points of an aqueous solution containing 39.5 grams of ethylene glycol (HOCH2CH2OH) dissolved in 750. mL of water.

Tb= [0.51K kg/mol] [(0.848m]

Tb=0.43 K therefore Tb= 100.43°C

Tf = [1.86 K kg/mol] [(0.848m]

Tf =1.58 K therefore Tf= -1.58°C

kind of slight… considerably more ethylene glycol needed for automobile use.