19012018

Question 1

Colligative properties

Answer 1

• Properties that depend on the no. of solute particles in solution
• NOT depend on the nature of solute particles
• Colligative properties are:
  1) Vapor-pressure lowering
  2) Boiling-point elevation
  3) Freezing-point depression
  4) Osmotic pressure (*)

Question 2

1) Vapor-pressure lowering (Raoult’s law)

Answer 2

Raoult’s law:
- The partial pressure of a solvent over a solution is given by the vapor pressure of the pure solvent times the mole fraction of the solvent in the solution
P_1 = x_1P_1^o,
where P_1 = partial pressure of solvent over solution, P_1^o = vapor pressure of pure solvent, x_1 = mole fraction of solvent
Change in P = x_2P_1^o,
where change in P = P_1^o - P_1, x_2 = mole fraction of solute

Question 3

1) Vapor-pressure lowering (volatile components)

Answer 3

• If both components of a solution are “volatile”, the vapor pressure of the solution is the sum of the individual partial pressures
  P_A = x_AP_A^o
  P_B = x_BP_B^o
  P_T = x_AP_A^o + x_BP_B^o

Question 4

1) Vapor-pressure lowering (ideal solution)

Answer 4

• An ideal solution obey Raoult’s law

Question 5

Example:
Calculate the vapor pressure of water over a solution made by dissolving 225 g of glucose in 575 g of water at 35C.
(At 35C, P_H2O^o = 42.2mmHg)

Answer 5

Step 1: figure out how many moles of glucose do we have
> 225 g glucose; molar mass of glucose = 180.2 g/mol
> no. of moles of glucose = 225 g glucose / 180.2 g/mol = 1.25 mol
Step 2: figure out the amount of water
> 575 g H2O / 18.02 g/mol = 31.9 mol H2O
Step 3: find out the mole fraction of water
> x_H2O = 31.9 mol H2O / (1.25 mol glucose + 31.9 mol H2O) = 0.962
Step 4: find out partial pressure of water
> P_H2O = x_H2OP_H2O^o = 0.96242.2mmHg = 40.6mmHg
(Make sense!! Reduction > should be less than the glucose solution)

Question 6

2) Boiling-point elevation

Answer 6

• Solutions boil at a higher temp than the pure solvent
  Change in T_b = T_b - T_b^o
  Change in T_b = K_b*m
  Where change in T_b = boiling point elevation, K_b = boiling point elevation constant (C/m), m = molality

Question 7

3) Freezing-point elevation

Answer 7

• Solutions freeze at a lower temperature than the pure solvent
  Change in T_f = T_f - T_f^o
  Change in T_f = K_f*m
  Where change in T_f = freezing point elevation, K_f = freezing point elevation constant (C/m), m = molality

Question 8

Example:
Ethylene glycol [CH2(OH)CH2(OH)] is a common automobile antifreeze. It is water soluble and fairly nonvolatile (b.p. 197 C). Calculate
A) Freezing point
B) Boiling point of a solution containing 685 g of ethylene glycol in 2075 g of water

Answer 8

Step 1: find out the number of mole of C2H6O2
685 g C2H6O2 / 62.07 g/mol = 11.04 mol C2H6O2
Step 2: find out molality (use kg!!!)
11.04 mol C2H6O2 / 2.075 kg H2O = 5.32 m C2H6O2
A) Change in T_f = K_fm = (1.86 C/m)(5.32 m) = 9.89 C
> Freezing point = (0 - 9.89 C) = -9.89 C
(new freezing point as a result of adding glycol)
B) Change in T_b = K_bm = (0.52 C/m)(5.32m) = 2.8 C
> Boiling point = (100 + 2.8 C) = 102.8 C

Question 9

Osmosis

Answer 9

• The selective passage of solvent molecules thru a porous membrane frm a more dilute solution to a more concentrated one