Answers to Problem set #12
1. When 1.2 g of sulfur is melted with 15 g of naphthalene, the solution freezes
at 77.2°C. What is the molar mass of this form of sulfur (data for
Naphthalene: melting point, 80°C; Kf = 6.8°C kg/mol)
ΔTf= Tf,soln-Tf,pure= -Kfmsolute
ΔTf = 77.2-80= -2.8°
m=molality= moles solute/kg solvent
15 g naphthalene = 0.015 kg naphthalene
m = (1.2g/ molar mass S)/ 0.015 kg napth
ΔTf =-2.8°C = -(6.8°C kg/mol) (1.2g/ molar mass S)/ 0.015 kg napth
(0.411mol/kg)(0.015 kg napth) = 1.2g/ molar mass S
molar mass S = 1.2 g / 0.006165 mol = 194.64 g/mol =S6
2. A solution of 5 g of unknown X dissolved in 20g benzene freezes at –4.52°C.
Benzene normally freezes at 5.48°C and its Kf is 5.12 °C kg/mol. What is
unknown X?
a. C6H4Cl2
b. C10H8 √
c. C10H20O
d. Some other compound
ΔTf= Tf,soln-Tf,pure= -Kfmsolute
ΔTf = -10 °C = -(5.12 °C kg/mol) [(5g/ molar mass X)/0.02 kg benzene]
1.95 mol/kg =(5g/ molar mass X)/0.02 kg benzene
molar mass X = 128 g/mol, C10 H8
3. What is the mass percent HBr (MW=80.91 g/mol) in an 11.4m solution of
HBr in water?
11.4 m = 11.4 moles HBr/ 1000g H2O = 922.37 g HBr / 1000g H2O
mass percent = mass HBr/mass HBr + mass H2O * 100%
mass percent= 922.37 g / 922 g + 1000g * 100% = 47.9 %
4. A solution of 0.5 g of an unknown compound creates an osmotic pressure of
4.7 mmHg when dissolved in one liter of water at 27°C. What is the
approximate molar mass of the compound?
Π = C RT (4.7 mmHg) (1atm/760 mmHg) =0.0062 atm = Π
C= (0.5 g/ molar mass) /1L; T= 300K; R= 0.0821 L atm/mol K
0.0062 atm = (0.5 g/ molar mass) /1L] (0.0821 L atm/mol K)(300K)
2.5x10-4 mol =(0.5 g/ molar mass)
molar mass = 2000 g/mol
5. Find the boiling point of a solution of 5.00 g of naphthalene (C10H8) in 100g of
benzene (Kb(benzene)= 2.53 °C/m; the normal boiling point of benzene
=80°C).
ΔTb= Tb,soln-Tb,pure= Kbmsolute
ΔTb= Tb,soln- 80°C= (2.53°C/m) (5g/128g/mol)/ 0.1 kg benzene
Tb,soln- 80°C= (2.53°C/m) * 0.039 mol napth/0.1 kg benzene
Tb,soln = 81°C
6. The freezing point of a 1 m aqueous solution of HF is found to be –1.91°C.
The freezing point depression constant for water is 1.86 °C kg/mol. What is
the percent of dissociation of HF at this concentration?
Colligative properties are a function of the number of particles
ΔTf= -Kfmsoluteisolute
ΔTf = -1.91°C = -(1.86°C/m) (1m) i
i=1.0268
i=1 for non-dissociable solutes
i=2 if HF completely dissociated.
HF(aq) <=> H+ (aq) + F- (aq)
Initial 1m
Eq. 1m-x
x
x
Total effective molality =1m-x +x +x = 1m+x = 1.0268; x=0.0268
fraction dissociated = x/1m = 0.0268. % dissociation = 2.7%
7. The vapor pressure of pure acetone (CH3COCH3) at 30°C is 0.327 atm.
Suppose 15 g of benzophenone, C13H10O is dissolved in 50 g of acetone
(C3H6O). Calculate the vapor pressure of acetone above the resulting solution.
P1 = X1 P1°; P1°=0.327 atm
Xacetone = (50 g / 58g/mol)/ [(50 g / 58g/mol)+ (15g / 182 g/mol)]=0.913
Pacetone = (0.913)( 0.327 atm) = 0.298 atm = 0.3 atm
8. An aqueous solution is 0.8402 molal in Na2SO4. It has a freezing point of
–4.214°C. Determine the effective number of particles arising from each
Na2SO4
ΔTf= -Kfmsoluteisolute ; ΔTf = –4.214°C; Kf (H2O) = 1.86 °C kg/mol
ΔTf = –4.214°C = -(1.86 °C/m) (0.8402 m)i
i= 2.69 ~ 3 particles per unit Na2SO4 dissolved.