1. Caffeine, C
8
H
10
N
4
O
2
, is a stimulant found in tea and coffee. A sample of the substance was dissolved in 45.0 g of chloroform, CHCl
3
, to give a 0.0946 m
solution. How many grams of caffeine were in the sample?
2. Concentrated aqueous ammonia contains 1.00 mol NH
3
dissolved in 2.44 mol H
2
O.
What is the mole fraction of NH
3
in concentrated aqueous ammonia? What is the
molal concentration of NH
3
?
3. A solution was prepared by dissolving 0.915 g of sulfur , S
8
, in 100.0 g of acetic
acid, HC
2
H
3
O
2
. Calculate the freezing point and boiling point of the solution.
4.
Butylated hydroxytoluene (BHT) is used as an antioxidant in processed foods. (It
prevents fats and oils from becoming rancid.) A solution of 2.500 g BHT in 100.0
g of benzene has a freezing point of 4.880 ° C. What is the molecular weight of
BHT?
5.
Assume that water and ethanol mix to form an ideal solution, i.e. their mixtures
obey Raoult's Law at all concentrations. The vapor pressure of pure water at 25°
C is 23.8 torr and the vapor pressure of pure ethanol at 25° C is 58.9 torr. For a
mixture of ethanol and water in which the mole fraction of ethanol is 0.200, what
is the vapor pressure of the solution and what is the mole fraction of ethanol in the
vapor?
6.
Ehylene glycol (HOCH
2
CH
2
OH) is a low volatility liquid commonly sold as antifreeze to be mixed with water in car radiators. By adding a nonvolatile solute to the water in the radiator, not only is the freezing point of the resulting solution substantially lower than the freezing point of water, the boiling point is raised substantially. Thus a water/ethylene glycol mixture protects the radiator both from freezing and from boiling over. Calculate the freezing and boiling points of a mixture of water and ethylene glycol obtained by mixing 2.00 kg ethylene glycol
(M.W. = 62.0682 g/mol) and 2.00 kg of water.
7.
A 1.14 g sample of a molecular substance dissolved in 100 g of camphor. The
resulting solution is observed to start freezing at a temperature of 177.3 ° C. What
is the molar mass of the solute?
8. Calculate the Osmotic pressure at 50 degrees Celsius of a glucose solution C
6
H
12
O
6
that has 60 grams of glucose dissolved in enough water to make 1500 ml.

1. Answer: molality = (moles solute)/(mass solvent)
So, moles solute = (molality)(mass solvent) = (0.0946 mol/kg)(0.045 kg) = 0.004257 mol. mass of solute = (moles)(molar mass) = (0.004257 mol)(194.194 g/mol) = 0.827 g
2. Answer: X
NH3
= (moles NH
3
)/(total moles) = (1.00 moles NH
3
)/(3.44 moles) = 0.291 mass of H
2
O = 2.44 mol(18.02 g/mol) = 43.97 g molality NH
3
= (moles NH
3
)/(mass solvent) = (1.00 mol)/(0.04397 kg) = 22.7 mol/kg
3. Answer: Calculate molality first: moles solute = 0.915 g(1 mol/ 256.528 g) =0.003567 mol molality = (moles solute)/(mass solvent) = (0.003567 mol)/(0.100 kg) = 0.03567 mol/kg
(1) T
T f f
= K f m = (3.59 ° C/m)(0.03567 m) = 0.1281° C
= T f
° - T f
= 16.60° C - 0.1281° C = 16.47° C
(2) T b
= K b m = (3.08 ° C/m)(0.03567 m) = 0.1099° C
T b
= T b
° + T b
= 118.5° C + 0.1099° C = 118.6° C
4. Answer: (1) T f
= T f
° - T f
= 5.455° C - 4.880° C = 0.575° C
(2) T f
= K f m so..... m = T f
/ K f
= (0.575° C)/(5.065 ° C/m) = 0.1135 mol/kg
(3) molality = (moles solute)/(mass solvent) so.... moles solute = (molality)(mass solvent)
= (0.1135 mol/kg)(0.100 kg) = 0.01135 mol
(4) molar mass = (mass)/(moles) = (2.500 g)/(0.01135 mol) = 220 g/mol
5 Answer. P(ethanol) = (.200)(58.9 torr) = 11.78 torr
P(water) = (.800)(23.8 torr) = 19.04 torr
P(total) = 11.78 torr + 19.04 torr = 30.82 torr +2 points
X vapor (ethanol) = (11.78 torr)/(30.82 torr) = 0.382
6. moles of ethylene glycol = (2000 g)(1 mol/62.0692 g) = 32.2 mol
molality = (moles solute)/(mass solvent) = (32.2 mol)/(2.00 kg) = 16.1 m
T b
= (16.1 m)(.512 ° C/m) = 8.25 ° C, so the boiling point is 108.25 ° C.
T f
= (16.1 m)(1.858 ° C/m) = 29.9 ° C, so the freezing point is -29.9 ° C.
7. (1) molality = T f
/K f
= (179.5 ° C - 177.3 ° C)/(40 ° C/m) = 0.055 m
(2) moles solute = (molality)(mass solvent) = (0.055 m)(0.100 kg) = 0.0055 m
(3) molar mass = mass/moles = (1.14 g)/(0.0055 mol) = 207g/mol +2 points
8.
1.
Convert grams of glucose to moles of glucose by dividing the mass by the molecular weight of glucose(180)
60 grams glucose X 1 mole glucose / 180 grams glucose = 0.333 moles glucose
2.
Convert milliliters of solution to liters
1500 ml X 1 liter / 1000 ml = 1.500 liter
3.
Determine the Molarity of the solution.
Molarity = 0.333 mole / 1.500 liter = 0.222 moles / liter
4.
Convert the Celsius temperature to Kelvin.
K = C + 273 = 50 + 273 = 323 K
5.
Calculate the Osmotic Pressure using the formula
Osmotic Pressure = (0.222) (0.0821) (323) = 5.89 atmospheres