1. vapor pressure curve for water

2. V. P. of the solution Mole Fraction of Solvent 01 V.P. of pure solvent V.P. (sol’n) =  (solv) V.P. (solv)

3. P T Normal F.P.Normal B.P. V.P. solid = V.P. liquid New, Lower F.P.  T = K f m 1.0 Atm V.P. liquid

4. What is the freezing point of an aqueous solution of MgSO 4 if 0.602 grams of magnesium sulfate was dissolved in 100. mL of water? (K f 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 T f = -0.0930°C m = 0.050 x 2  T = K f mi van’t Hoff Factor 2  T=0.186 K therefore T f = -0.186°C ….BUT WAIT!!

5. P T New, higher B.P.Normal B.P. V.P. solid = V.P. liquid  T = K b mi 1.0 Atm V.P. liquid

6. What is the boiling point of an aqueous solution of MgSO 4 if 0.602 grams of magnesium sulfate was dissolved in 100. mL of water? (K b for water is 0.51 K kg/mol)  T= [0.51K kg/mol] [(0.0500m] 2  T=0.051 K therefore T b = 100.051°C  T (meas.) =0.033 K therefore T b (meas.) = 100.033°C Therefore, the i value must only be 1.3; how come? ….BUT WAIT!!

7. What is the freezing point of an aqueous solution of MgSO 4 if 0.602 grams of magnesium sulfate was dissolved in 100. mL of water? (K f 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 T f = -0.0930°C….BUT WAIT!! m = 0.050 x 2  T = K f mi van’t Hoff Factor 2  T=0.186 K therefore T f = -0.186°C x 1.3 1.3 0.121 K -0.121° C

8. Have you ever heard of ethylene glycol? Calculate the boiling and freezing points of an aqueous solution containing 39.5 grams of ethylene glycol (HOCH 2 CH 2 OH) dissolved in 750. mL of water.  T b = [0.51K kg/mol] [(0.848m]  T b =0.43 K therefore T b = 100.43°C  T f = [1.86 K kg/mol] [(0.848m]  T f =1.58 K therefore T f = -1.58°C kind of slight… considerably more ethylene glycol needed for automobile use.