CHEM1909
2004-N-6
November 2004
 Sucrose, C12H22O11, is dissolved in 100 mL of water at 25 °C. The osmotic pressure
is measured as 12.2 atm. What is the mass of sucrose that was originally dissolved?
The osmotic pressure, Π, is given by Π = MRT where M is the molarity of the
solution. Hence,
M=
Π
(12.2atm)

 0.50M
RT (0.08206Latm K 1 mol 1 )  ((25  273)K)
(As the pressure is given in units of atm, R = 0.08206 L atm K-1 mol-1 has been
used, leading to molarity in units of mol L-1 or M).
As concentration =
number of moles
, the amount present in 100 mL is:
volume
n = concentration × volume = (0.50 M) × (100 × 10-3 L) = 0.050 mol
The molar mass of sucrose is:
molar mass = (12 × 12.01 (C)) + (22 × 1.008 (H)) + (11 × 16.00 (O)) g mol-1
= 342.296 g mol1
Hence, this amount of sucrose corresponds to a mass of:
mass = number of moles × molar mass = (0.050 mol) × (342.296 g mol-1)
= 17 g
Answer: 17 g
ANSWER CONTIUNES ON THE NEXT PAGE
Marks
8
CHEM1909
2004-N-6
November 2004
Calculate the vapour pressure of water above this solution, given that at 25 °C, the
density of water is 0.997 g mL–1 and the vapour pressure of water is 23.8 mmHg.
o
According to Raoult’s Law, the vapour pressure of pure solvent, Psolvent
is
o
lowered by the presence of solute by an amount ΔP where ΔP = Xsolute × Psolvent
where Xsolute is the mole fraction of the solute.
From above, 0.050 mol of sucrose are present in 100 mL of water. Using the
density, this volume of water corresponds to a mass of:
mass = density × volume = (0.997 g mL-1) × (100 mL) = 99.7 g
The molar mass of H2O is (2 × 1.008 (H)) + 16.00 (O) g mol1 = 18.016 g mol1 so
the number of moles of water is:
moles =
mass
99.7 g

 5.53mol
molar mass 18.016g mol 1
Hence, the mole fraction of the solute, sucrose, is:
Xsolute =
moles of solute
0.050

 8.98  103
moles of solute  moles of solvent (0.050  5.53)
Thus,
o
ΔP = Xsolute × Psolvent
= (8.98 × 10-3) × 23.8 mmHg = 0.213 mmHg
The vapour pressure is lowered to (23.8 – 0.213) mmHg = 23.6 mmHg.
As 760 mmHg = 1 atm, this corresponds to
23.6
 0.0311atm
760
Answer: 23.6 mmHg or 0.0311 atm
ANSWER CONTIUNES ON THE NEXT PAGE
CHEM1909
2004-N-6
November 2004
If this solution is heated at 1.00 atm, at what temperature will the water boil?
The molal boiling point elevation constant (Kb) for pure water is 0.512 °C kg mol–1.
From above, 0.050 of sucrose are dissolved in 99.7 g of water. The molality is:
molality =
number of moles of solute(mol)
0.050mol

 0.50mol kg 1
3
mass of solvent (kg)
(99.7  10 kg)
The boiling point elevation, ΔTb, is given by:
ΔTb = Kbm
where Kb is the molal boiling point elevation constant and m is the molality.
Hence,
ΔTb = Kbm = (0.512 °C kg mol-1) × (0.50 mol kg-1) = 0.26 °C
At atmospheric pressure, the boiling point of water is 100 °C. The solution will
boil at 100.26 °C.
Answer: 100.26 °C
CHEM1909
2005-N-5
November 2005
 Consider the following equilibrium in the gas-phase at 35 ºC.
Cl2(g) + 2NO(g)
2NOCl(g)
Kc = 6.25  104 M–1
Equimolar amounts of NOCl(g) and Cl2(g) are introduced into a sealed 1.00 L flask.
When the system reaches equilibrium at 35 ºC, the concentration of NO(g) in the
flask is 4.04  10–4 M. What amount of Cl2(g) (in mol) was initially added to the
flask?
If the initial concentrations of Cl2(g) and NOCl (g) are both equal to x, the
reaction table is:
Cl2(g)
2NO(g)
2NOCl(g)
initial
x
0
x
change
+y
+2y
-2y
equilibrium
x+y
2y
x – 2y
(As [NO(g)] has increased so must [Cl2(g)] but by half as much since every two
NO molecules are made for every Cl2. [NOCl(g)] must have decreased by the
same amount as [NO(g)] has increased. The reaction is proceeding backwards
from the point at which it has begun.)
As [NO(g)]equilibrium = 2y = 4.04 × 10-4 M, y = 2.02 × 10-4 M and so:
[Cl2(g)]equilibrium = x + y = x + 2.02 × 10-4 M and
[NOCl(g)]equilibrium = x - 2y = x – 4.04 × 10-4 M
From the equilibrium constant:
Kc 
[NOCl(g)]2
( x  4.04 x104 )2

[Cl 2 (g)][NO(g)]2 ( x  2.02 x104 )(4.04 x104 )2
As the equilibrium for the forward reaction is very large (6.25 x 104), the
amount of Cl2 and NO that are produced in the backward reaction is very
small. It can be assumed that this amount (4.04 x 10-4) is small compared to x.
Hence, x + 2.02 x 10-4 ~ x and x – 4.04 x 10-4 ~ x:
Kc 
( x  4.04 x104 )2
x2
x
~

4
4 2
4 2
( x  2.02 x10 )(4.04 x10 )
x(4.04 x10 )
(4.04 x10 4 ) 2
As Kc = 6.25 × 104, [Cl2(g)]initial = x = 0.0102 M. As the volume of the flask is 1.00
L, the number of moles = concentration × volume = 0.0102 × 1.00 = 0.0102 mol
Answer: 0.0102 M
ANSWER CONTINUES ON THE NEXT PAGE
Marks
6
CHEM1909
2005-N-5
November 2005
At the same temperature (35 ºC) O2(g) reacts with NO(g) according to the equation:
O2(g) + 2NO(g)
Kc = 6.25 M–1
2NO2(g)
Determine Kc for the following reaction.
2NO2(g) + Cl2(g)
2NOCl(g) + O2(g)
The equilibrium constants for the three reactions are:
[NO2 ]2
[NOCl]2 [O2 ]
[NOCl]2
, K c (2)=
and K c (3)=
K c (1)=
[Cl 2 ][NO]2
[O2 ][NO]2
[NO2 ]2 [Cl 2 ]
so that
K c (1)
[NO2 ]2
[NOCl]2 [O2 ]
[NOCl]2
=
/
=
= K c (3)
K c (2) [Cl 2 ][NO]2 [O2 ][NO]2 [NO2 ]2 [Cl 2 ]
6.25  104
Hence, Kc (3) =
= 1.00 × 104.
6.25
Kc = 1.00 × 10-4 M
ANSWER CONTINUES ON THE NEXT PAGE
CHEM1909
2005-N-6
November 2005
Calculate the partial pressure equilibrium constant, Kp, at 35 ºC for the reaction:
2NO2(g) + Cl2(g)
2NOCl(g) + O2(g)
In the reaction, 3 moles of gas react to form 3 moles of gas – there is no
change in the number of moles of gas and so Δn = 0. Kc was calculated in the
previous question (2005-N-5) to be 1.00 × 10-4
K p = K c (RT )n
= K c ( RT )0 = K c  1.00×104
Kp = 1.00 × 10-4
What is the standard free energy change, ∆G, for the forward reaction (in kJ mol–1)
at 35 ºC?
Using ΔG° = -RTlnKp,
ΔG° = -(8.314 J K-1 mol-1) × ((35 + 273) K) × ln(1.00 × 10-4)
= -23600 J mol-1 = -23.6 kJ mol-1
G = -23.6 kJ mol-1
If 0.150 mol of O2(g) and 3.00  10–4 mol of NO2(g) are added to the 1.00 L flask,
determine the free energy change, ∆G, (in kJ mol–1) as the system moves to its new
equilibrium point.
From the previous question (2005-N-5), the initial concentrations of [NOCl] and [Cl2]
are both equal to 0.0102 M. The concentrations of O2(g) and NO2(g) are equal to
0.150 M and 3.00 × 10-4 M as the flask has a volume of 1.00 L.
[NOCl(g)]2 [O2 (g)]
The reaction quotient, Q, for this reaction is Q =
. Hence:
[NO2 (g)]2 [Cl 2 (g)]
Q=
(0.0102)2 (0.150)
= 17000 and
(3.00  10-4 )2 (0.0102)
ΔG = ΔG o +RTlnQ =(-23.6  103 J mol -1 )+(8.314 J K -1 mol -1 )  ((35+273) K)  ln(17000)
= +1.4kJ mol -1
G = +1.4 kJ mol-1
Will the amount of NO2(g) in the flask increase or decrease as the system moves to its
new equilibrium position? Explain.
As Q > Kc, the system will move to decrease Q. This occurs by reducing the
concentration of the products and increasing the concentration of reactants.
[NO2(g)] will thus increase.
Marks
7
CHEM1909
2005-N-7
November 2005
 Calcium chloride (3.42 g) is completely dissolved in 200 mL of water at 25.00 ºC in a
‘coffee cup’ calorimeter. The temperature of the water after dissolution is 27.95 ºC.
Calculate the standard enthalpy of solution of CaCl2 (in kJ mol–1). The heat capacity
of water is 4.184 J K–1 g–1. Ignore the heat capacity of the CaCl2.
As water has a density of 0.997 g mL-1, 200 mL of water corresponds to a mass
of:
m = density × volume = (0.997 g mL-1) × (200 mL) = 199 g
The temperature change, ΔT, is (27.95 – 25.00) ºC = 2.95 ºC (or 2.95 K)
The heat change accompanying a change in temperature is given by:
q = m × C × ΔT = (199 g) × (4.184 J K-1 g-1) × (2.95 K) = 2460 J = 2.46 kJ.
As the temperature of the water increases, the dissolution is exothermic and
hence, ΔH = -2.46 kJ.
The formula mass of CaCl2 is 40.08 (Ca) + (2 × 35.45 (Cl)) g mol-1 = 110.98 g
mol-1. Hence, 3.42 g corresponds to:
number of moles =
mass
3.42g

 0.0308 mol
formulamass 110.98g mol 1
If this amount gives rise to an heat change of 2.46 kJ, the molar enthalpy change
2.46kJ
is given by
 79.9kJ mol 1
0.0308mol
Answer: -79.9 kJ mol-1
ANSWER CONTIUNES ON THE NEXT PAGE
Marks
8
CHEM1909
2005-N-7
November 2005
What would be the vapour pressure of water above this solution?
(P0 (H2O) = 3.17 kPa)
o
According to Raoult’s Law, the vapour pressure of pure solvent, Psolvent
is
lowered by the presence of solute by an amount ΔP:
o
ΔP = i × Xsolute × Psolvent
where Xsolute is the mole fraction of the solute and i is the amount (mol) of
particles in solution divided by the amount (mol) of dissolved solute.
The molar mass of H2O is (2 × 1.008 (H)) + 16.00 (O) g mol-1 = 18.016 g mol1 so
the number of moles of water in 199 g is:
moles =
mass
199g

 36.0mol
molar mass 18.016g mol 1
As the solution contains 0.0308 mol of CaCl2, the mole fraction of the solute is:
Xsolute =
moles of solute
0.0308

 8.55  104
moles of solute  moles of solvent (0.0308  36.0)
CaCl2 dissolves to give three particles, Ca2+ + 2Cl-, so i = 3. Thus,
o
ΔP = i × Xsolute × Psolvent
= 3 × (8.55 × 10-4) × 3.17 kPa = 8.14 × 10-3 kPa
The vapour pressure is lowered to (3.17 – 8.14 × 10-3) kPa = 3.16 kPa.
Answer: 3.16 kPa
ANSWER CONTIUNES ON THE NEXT PAGE
CHEM1909
2005-N-7
November 2005
What would be the freezing point of this solution? The molal freezing point
depression constant (Kf) for water is 1.86 °C kg mol–1.
From above, 0.0308 mol of CaCl2 are dissolved in 199 g of water. The molality
is:
molality =
number of moles of solute(mol)
0.0308mol

 0.155mol kg 1
3
mass of solvent (kg)
(199  10 kg)
The freezing point depression, ΔTf, is given by:
ΔTf = i × Kfm
where Kf is the molal freezing point depression constant and m is the molality.
As i = 3,
ΔTf = i × Kfm = 3 × (1.86 °C kg mol–1) × (0.155 mol kg-1) = 0.865 °C
At atmospheric pressure, the water freezes at 0 °C. The solution will freeze
at -0.865 °C.
Answer: -0.865 °C
Which would you expect to cause the greater freezing point depression of water,
3.42 g of CaCl2 or 3.42 g of NaCl? Explain your answer.
The formula mass of NaCl is 22.99 (Na) + 35.43 (Cl) g mol-1 = 58.42 g mol-1. The
number of moles of NaCl is therefore:
number of moles =
mass
3.42g

 0.0585mol
formulamass 58.42g mol 1
NaCl dissolves to give two particles, Na+ + Cl- so i = 2.
As ΔTf = i × Kfm, the freezing point depression of x kg of water is given by:
ΔTf = i × Kfm = i × Kf ×
For NaCl, ΔTf = 2 × Kf ×
moles of solute
x
0.0585
0.0308
. For CaCl2, ΔTf = 3 × Kf ×
.
x
x
Hence, ΔTf α 0.117 for NaCl and ΔTf α 0.0924 for CaCl2. The freezing point
depression is larger for NaCl.