1. Hydrogen peroxide (H2O2) is used in concentrated solution in rocket fuels and dilute solution as hair bleach. An aqueous
solution of H2O2 is 30.0% by mass and has a density of 1.11 g/mL. Calculate the following quantities for this solution:
(a) molality of H2O2, (b) mole fraction of H2O2, and (c) molarity of H2O2.
Assume a 100.0 g solution which means that 30.0 g H2O2 and 70.0 g H2O are present in the solution.
(a) molality of H2O2
molality = moles of H2O2/kg H2O
(30.0 g H2O2)(1 mol H2O2/34.01 g H2O2) = 0.882 mol H2O2
70.0 g is the same as 0.0700 kg
molality = 0.882 mol/0.0700 kg = 12.6 m
(b) mole fraction of H2O2
(70.0 g H2O)(1 mol H2O/18.015 g) = 3.89 mol H2O
X of H2O2 = 0.882 mol/(0.882 mol + 3.89 mol) = 0.882 mol/4.77 mol
X of H2O2 = 0.185
(c) molarity of H2O2
volume = mass/density
(100. g)(1.00 mL/1.11 g) = 90.1 mL = 0.0901 L
molarity = moles of H2O2/ L solution = 0.882 mol H2O2/ 0.0901 L = 9.79 M
2. Which solvent, water or carbon tetrachloride (CCl4), would you choose to dissolve each of the following compounds?
Justify your reasoning.
(a) LiBr
LiBr is an ionic compound which dissolves in water because of ion-dipole attractive forces.
(b) CH3OH
CH3OH is a polar compound which dissolves in water because of hydrogen bonding interactions.
(c) CH3(CH2)8CH2OH
This compound has a long carbon chain which determines its solubility properties. This carbon chain is hydrophobic and so this compound dissolves in carbon tetrachloride because of dispersion forces.
3. (a) Which compound in each of the following pairs do you predict to be more soluble in water? Justify your reasoning in
each case.
1. toluene (C7H8) or diethyl ether (CH3CH2OCH2CH3)
Toluene is essentially a nonpolar molecule whereas diethyl ether is a polar molecule. We would expect diethyl ether to be more soluble in water because of hydrogen bonding interactions with water.
2. chloroform (CHCl3) or acetone (CH3COCH3)
Even though both of these molecules are polar, we would expect acetone to be more soluble in water because of hydrogen bonding interactions with water.
(b) Which compound in each of the following pairs do you predict to be more soluble in benzene (C6H6)? Justify your
reasoning in each case.
1. cyclohexane (C6H12) or methanol (CH3OH)
Cyclohexane is essentially a nonpolar molecule whereas methanol is a polar molecule. Since benzene is a nonpolar molecule we would expect cyclohexane to be more soluble in benzene due to dispersion forces.
2. I2 or MgCl2
I2 is a nonpolar molecule whereas MgCl2 is an ionic compound. Since benzene is a nonpolar molecule we would expect I2 to be more soluble in benzene due to dispersion forces.
4. (a) Calculate the concentration of CO2 in a soft drink that is bottled with a partial pressure of CO2 of 4.0 atm over the
liquid at 25 °C. The Henry’s Law constant for CO2 in water at this temperature is 3.4 x 10 – 2 M/atm.
Sgas = kHPgas
Sgas = (3.4 x 10 – 2 M/atm)(4.0 atm) = 0.14 M
(b) Calculate the concentration of CO2 in a soft drink after the container is opened and equilibrates at 25°C under a CO2
partial pressure of 3.0 x 10 – 4 atm.
Sgas = kHPgas
Sgas = (3.4 x 10 – 2 M/atm)(3.0 x 10 – 4 atm) = 1.0 x 10 – 5 M
5. Calculate the boiling point and freezing point of a 25.0 % by mass solution of ethylene glycol (C2H6O2) in water. See
Table 12.8 for the values of Kb, Kf, Tb, and Tf for water.
We need to determine the molality of the solution.
molality = moles of C2H6O2/kg H2O
If we assume we have 100.0 g of this solution then we have 25.0 g of ethylene glycol and 75.0 g of water.
molar mass of ethylene glycol = 62.07 g/mol
moles = (25.0 g C2H6O2)(1 mol/62.07 g C2H6O2) = 0.403 mol C2H6O2
molality = (0.403 mol C2H6O2)/(0.0750 kg H2O) = 5.37 m
We can now calculate the boiling point elevation and freezing point depression of this solution.
DTb = mKb = (5.37 m)(0.512 °C/m) = 2.75 °C
DTf = mKf = (5.37 m)(1.86 °C/m) = 9.99 °C
The boiling point of the solution = 100.00 °C + 2.75 °C = 102.75 °C
The freezing point of the solution = 0.00 °C – 9.99 °C = – 9.99 °C
6. To determine the molar mass of the anti-freeze protein from the Arctic right-eye flounder, the osmotic pressure of a
solution containing 13.2 mg of protein per mL of solution was measured and found to be 21.2 torr at 10°C. What is the
molar mass of the protein?
We must first convert mass, volume, pressure, and temperature to the proper units.
mass = 13.2 mg x (1 g/1000 mg) = 0.0132 g
volume = 1.00 mL x (1 L/1000 mL) = 0.00100 L
pressure = 21.2 torr x (1 atm/760 torr) = 0.0279 atm
temperature = 10°C + 273 = 283 K
P = MRT Þ M = P /(RT)
M = 0.0279 atm/[(0.08206 L atm/mol K)(283 K)]
M = 1.20 x 10-3 mol/L
moles = molarity x volume in liters
moles = (1.20 x 10-3 mol/L) x (0.00100 L) = 1.20 x 10-6 mol
molar mass = mass/mol = 0.0132 g/(1.20 x 10-6 mol)
molar mass = 1.10 x 104 g/mol
