Chem 020 Tutorial Test 4, Section 001
Tuesday November 13, 07/08
Name:
TA:
1. Bromochloromethane, ClCH2Br, has a molar mass of 129.39 g/mol. Its normal
boiling point is 68.11 C when its vapor pressure is 101.3 kPa. Find the density of
bromochloromethane in the vapor phase at its boiling point.
From PV = nRT it is easy to show:
M = dRT/P so that: 129.39 = ( d x 8.314 x (273 + 68.11))/101.39
d = 4.63 g/L
1 mark for the formula 1 for the correct calculation.
2. Two flasks are connected by a stopcock both flasks are maintained at a
temperature of 30 C. The first, with a volume of 100 liters contains Ar at a
pressure of 50 kPa, the second with a volume of 50 liters contains Xe at a
pressure of 75 kPa. If the stopcock is opened and the gases are allowed to mix
without any temperature change. Find the average molar mass of the resulting gas
mixture.
First calculate the new pressure for each gas:
Ar P1V1 = P2V2 = 50 x 100 = P2 x 150(new volume) P2 = 33.33 kPa = PAr
Xe P1V1 = P2V2 = 75 x 50 = P2 x 150(new volume) P2 = 25 kPa = PXe
New PT = PAr + PXe = 33.33 + 25 = 58.33 kPa
Now PAr = XAr x PT = XAr x 58.33 = 33.33 so XAr = 0.57 XXe = 1 – 0.57 = 0.43
M (average) = XAr x molar mass Ar
0.43 x 131.3 = 79.2g
+
XXe x molar mass Xe = 0.57 x 39.95 +
2 marks for finding new total pressure, 1 for finding the average molar mass.
3. a) How many mL of an impure liquid CS2 mixture (density 1.26 g/mL) which
contains 85% of CS2 by mass must be burned to obtain 12.8 g of SO2 ?
b) If the resulting SO2 is collected at 25 C and 90 kPa, what volume of SO2 is
produced?
a) Reaction is:
CS2 + 3 O2 → CO2 + 2 SO2
Moles SO2 produced = mass/molar mass = 12.8/64 = 0.2 moles
Moles CS2 required = 0.2/2 = 0.1
Mass CS2 = moles x molar mass = 0.1 x 76 = 7.6 g
But [( mass CS2 )/ (mass liquid)] x 100 = 85 = [(7.6)/(mass liquid)] x 100
Mass liquid = 8.94
Density of liquid = 1.26g/mL = mass liquid/ volume of liquid = 8.94/ volume of
liquid
Volume liquid = 7.1 mL
b) PV = nRT V = nRT/P = (0.2 x 8.314 x (25+ 273))/90 = 5.5 L
1mark for the balanced equation, 1mark for volume of liquid, 1 mark for volume of
SO2 produced
4. Consider the reaction:
2 CO(g)
+
O2(g)
→ 2 CO2(g)
If the partial pressure of CO before reaction is 75 kPa and that of O2 is 175 kPa, find
the total pressure after reaction assuming the temperature and volume are constant.
After reaction Pressure of CO2 = Pressure of CO(g) before reaction = 75 kPa
After reaction Pressure of O2 = Pressure O2 before – 1/2 (pressure of CO(g) before
reaction) = 175 kPa 75/2 kPa = 137.5
P Total after reaction = P of CO2 after reaction + P of O2 after reaction = 75 + 137.5
= 212.5 kPa
2 marks
Chem 020 Tutorial Test 4, Section 004
Wednesday November 14, 07/08
Name:
TA:
1. Bromochloromethane, ClCH2Br, has a molar mass of 129.39 g/mol. Its normal
boiling point is 68.11 C when its vapor pressure is 101.3 kPa. Given that the
density of liquid ClCH2Br at 68.11 C is 1.12 g/ml, calculate the factor by which
it’s volume increases when it enters the vapor phase at this temperature.
Want : V(g)/ V(l)
It is easiest to assume 1.12 g of ClCH2Br are vaporized, since we know the
volume of the liquid from the density, V(l) = 0.001 L
Can now get V(g) from: PV = nRT
n = mass/MM= 1.12/129.39 = 0.008656
V(g) = (0.008656 x 8.314 x(68.11 + 273))/101.3 = 0.242
V(g)/ V(l) = .242/.001 = 242
2 marks
2. Two flasks are connected by a stopcock both flasks are maintained at a
temperature of 30 C. The first, with a volume of 50 liters contains Ar at a pressure
of 50 kPa, the second with a volume of 20 liters contains Xe at a pressure of 75
kPa. If the stopcock is opened and the gases are allowed to mix without any
temperature change. Find the average molar mass of the resulting gas mixture.
First calculate the new pressure for each gas:
Ar P1V1 = P2V2 = 50 x 50 = P2 x 70(new volume) P2 = 35.71 kPa = PAr
Xe P1V1 = P2V2 = 75 x 20 = P2 x 70(new volume) P2 = 21.42 kPa = PXe
New PT = PAr + PXe = 35.71 + 21.42 = 57.13 kPa
Now PAr = XAr x PT = XAr x 57.13 = 35.71 so XAr = 0.625 XXe = 1 – 0.625 =
0.375
M (average) = XAr x molar mass Ar
0.375 x 131.3 = 74.2g
+
XXe x molar mass Xe = 0.625 x 39.95 +
2 marks for finding new total pressure, 1 for finding the average molar mass.
3. a) How many mL of an impure liquid CS2 mixture (density 1.26 g/mL) which
contains 55% of CS2 by mass must be burned to obtain 12.8 g of SO2 ?
b) If the resulting SO2 is collected at 25 C and 90 kPa, what volume of SO2 is
produced?
a) Reaction is:
CS2 + 3 O2 → CO2 + 2 SO2
Moles SO2 produced = mass/molar mass = 12.8/64 = 0.2 moles
Moles CS2 required = 0.2/2 = 0.1
Mass CS2 = moles x molar mass = 0.1 x 76 = 7.6 g
But [( mass CS2 )/ (mass liquid)] x 100 = 55 = [(7.6)/(mass liquid)] x 100
Mass liquid = 13.82g
Density of liquid = 1.26g/mL = mass liquid/ volume of liquid = 13.82/ volume of
liquid
Volume liquid = 10.97 mL
b) PV = nRT V = nRT/P = (0.2 x 8.314 x (25+ 273))/90 = 5.5 L
1mark for the balanced equation, 1mark for volume of liquid, 1 mark for volume of
SO2 produced
4. Consider the reaction:
2 CO(g)
+
O2(g)
→ 2 CO2(g)
If the partial pressure of CO before reaction is 175 kPa and that of O2 is 175 kPa,
find the total pressure after reaction assuming the temperature and volume are
constant.
After reaction Pressure of CO2 = Pressure of CO(g) before reaction = 75 kPa
After reaction Pressure of O2 = Pressure O2 before – 1/2 (pressure of CO(g) before
reaction) = 175 kPa -175/2 kPa = 87.5
P Total after reaction = P of CO2 after reaction + P of O2 after reaction = 75 + 87.5 =
162.5 kPa
2 marks
Chem 020 Tutorial Test 4, Section 004
Monday November 19, 07/08
Name:
TA:
1. Bromochloromethane, ClCH2Br, has a molar mass of 129.39 g/mol. Its normal
boiling point is 68.11 C when its vapor pressure is 101.3 kPa. Calculate the
density of Bromochloromethane in the vapor phase at its normal boiling point.
From; PV = nRT it is easy to show that:
Molar Mass = dRT/P
Whence 129.39 = d x ((8.314 x (68.11 + 273))/101.3
d = 4.62 g/L
2
marks
2. Two flasks are connected by a stopcock both flasks are maintained at a
temperature of 30 C. The first, with a volume of 50 liters contains Ar at a pressure of
50 kPa, the second with a volume of 20 liters contains N2 at a pressure of 75 kPa. If
the stopcock is opened and the gases are allowed to mix without any temperature
change. Find the average molar mass of the resulting gas mixture.
First calculate the new pressure for each gas:
Ar P1V1 = P2V2 = 50 x 50 = P2 x 70(new volume) P2 = 35.71 kPa = PAr
N2 P1V1 = P2V2 = 75 x 20 = P2 x 70(new volume) P2 = 21.42 kPa = PN2
New PT = PAr + PN2 = 35.71 + 21.42 = 57.13 kPa
Now PAr = XAr x PT = XAr x 57.13 = 35.71 so XAr = 0.625 XXe = 1 – 0.625 =
0.375
M (average) = XAr x molar mass Ar
0.375 x 28 = 35.47g
+
XN2 x molar mass N2 = 0.625 x 39.95 +
2 marks for finding new total pressure, 1 for finding the average molar mass.
3. a) 13.82g of an impure liquid CS2 mixture (density 1.26 g/mL) were burned to
yield 12.8 g of SO2. Assuming that the CS2 is the only source of SO2 and that
combustion is complete, calculate the mass % of CS2 in the original mixture.
b) If the resulting SO2 is collected at 25 C and 90 kPa, what volume of SO2 is
produced?
a) Reaction is:
CS2 + 3 O2 → CO2 + 2 SO2
Moles SO2 produced = mass/molar mass = 12.8/64 = 0.2 moles
Moles CS2 required = 0.2/2 = 0.1
Mass CS2 = moles x molar mass = 0.1 x 76 = 7.6 g
% CS2 = (mass CS2/ Total Mass sample) x 100 = (7.6/13.82) x 100 = 55 %
b) PV = nRT V = nRT/P = (0.2 x 8.314 x (25+ 273))/90 = 5.5 L
1 mark for the balanced equation, 1mark for volume of liquid, 1 mark for volume
of SO2 produced
4. Consider the reaction:
2 SO2(g)
+
O2(g)
→ 2 SO3(g)
If the partial pressure of SO2 before reaction is 100 kPa and that of O2 is 175 kPa,
find the total pressure after reaction assuming the temperature and volume are
constant.
After reaction Pressure of SO3 = Pressure of SO2(g) before reaction = 50 kPa
After reaction Pressure of O2 = Pressure O2 before – 1/2 (pressure of SO2(g) before
reaction) = 175kPa -100/2 kPa = 125 kPa
P Total after reaction = P of SO3 after reaction + P of O2 after reaction = 100 + 125 =
225 kPa
2 marks
Chem 020 Tutorial Test 4, Section 001
Tuesday November 20, 07/08
Name:
TA:
1. A mixture of H2 and He at 300 K and 101.3 kPa has a density of 0.156 g/L.
Find the average molar mass of the mixture and the mole fraction of He in the
mixture.
From; PV = nRT it is easy to show that:
Molar Mass = dRT/P
Whence average molar mass = (0.156 x 8.314 x 300)/101.3 = 3.84 g
And
Average molar mass = (mole fraction H2 x Molar mass H2) + (mole fraction He x
molar mass He)
Let mole fraction He = x, then mole fraction H2 = 1-x
And 3.84 = (1-x) 2 + 4 x
x = mole fraction He = 0.92
1
mark for average molar mass 1 mark for mole fraction He
2. Which of the following statements are correct for TWO moles of an ideal gas?
Put a check at the end of each true statement.
A plot of P versus T at constant V will give a straight line.
A plot of V versus T at constant P will give a straight line with slope = R/P
A plot of PV/T versus P will give a straight line having a zero slope.
A plot of PV versus T will give a straight line having a zero slope.
A plot of PV/RT versus P will give a straight line having a slope = 2.
Two marks but only for the tow correct answers, if an additional incorrect response is
indicated deduct 1.
3. a) 13.82g of an impure liquid CS2 mixture (density 1.26 g/mL) were burned to
yield 12.8 g of SO2. Assuming that the CS2 is the only source of SO2 and that
combustion is complete, calculate the mass % of CS2 in the original mixture.
b) If the resulting SO2 is collected at 25 C and 90 kPa, what volume of SO2 is
produced?
a) Reaction is:
CS2 + 3 O2 → CO2 + 2 SO2
Moles SO2 produced = mass/molar mass = 12.8/64 = 0.2 moles
Moles CS2 required = 0.2/2 = 0.1
Mass CS2 = moles x molar mass = 0.1 x 76 = 7.6 g
% CS2 = (mass CS2/ Total Mass sample) x 100 = (7.6/13.82) x 100 = 55 %
b) PV = nRT V = nRT/P = (0.2 x 8.314 x (25+ 273))/90 = 5.5 L
1 mark for the balanced equation, 1mark for volume of liquid, 1 mark for volume
of SO2 produced Total = 3 marks
4. A vacuum chamber contains gas exerting a pressure of 1.33 x 10-10 kPa.
Find the number of gas molecules in one liter at this pressure, given the temperature
is 273 K.
Assume V = one liter
n = PV/RT = (1.33 x 10-10 x 1.0)/(8.314 x 273) = 5.86 x 10-14
# of molecules per liter = 5.86 x 10-14 x 6.02 x 1023 = 3.53 x1010 molecules
2 marks
5. If the number of moles and the temperature remain constant, by what factor must
the pressure be changed to double the volume occupied by the gas.
Pressure must be reduced by a factor of 2 (or ½ original pressure) accept both.
One mark.
Chem 020 Tutorial Test 4, Section 004
Wednesday November 21, 07/08
Name:
TA:
a.
Bromochloromethane, ClCH2Br, has a molar mass of 129.39 g/mol. Its
normal boiling point is 68.11 C when its vapor pressure is 101.3 kPa. Given that the
density of liquid ClCH2Br at 68.11 C is 1.12 g/ml, calculate the factor by which it’s
volume increases when it enters the vapor phase at this temperature.
Want : V(g)/ V(l)
It is easiest to assume 1.12 g of ClCH2Br are vaporized, since we know the
volume of the liquid from the density, V(l) = 0.001 L
Can now get V(g) from: PV = nRT
n = mass/MM= 1.12/129.39 = 0.008656
V(g) = (0.008656 x 8.314 x(68.11 + 273))/101.3 = 0.242
V(g)/ V(l) = .242/.001 = 242
2
1.
marks.
A mixture of H2 and He at 300 K and 101.3 kPa has a density of 0.156 g/L.
Find the average molar mass of the mixture and the mole fraction of He in the
mixture.
From; PV = nRT it is easy to show that:
Molar Mass = dRT/P
Whence average molar mass = (0.156 x 8.314 x 300)/101.3 = 3.84 g
And
Average molar mass = (mole fraction H2 x Molar mass H2) + (mole fraction He x
molar mass He)
Let mole fraction He = x, then mole fraction H2 = 1-x
And 3.84 = (1-x) 2 + 4 x
x = mole fraction He = 0.92
2
marks for average molar mass 1 mark for mole fraction He
2. Which of the following statements are correct for TWO moles of an ideal gas?
Put a check at the end of each true statement.
A plot of P versus T at constant V will give a straight line.
A plot of V versus T at constant P will give a straight line with slope = R/P
A plot of PV/T versus P will give a straight line having a zero slope.
A plot of PV versus T will give a straight line having a zero slope.
A plot of PV/RT versus P will give a straight line having a slope = 2.
Two marks
4. a) 13.82g of an impure liquid CS2 mixture (density 1.26 g/mL) were burned to
yield 12.8 g of SO2. Assuming that the CS2 is the only source of SO2 and that
combustion is complete, calculate the mass % of CS2 in the original mixture.
b) If the resulting SO2 is collected at 25 C and 90 kPa, what volume of SO2 is
produced?
a) Reaction is:
CS2 + 3 O2 → CO2 + 2 SO2
Moles SO2 produced = mass/molar mass = 12.8/64 = 0.2 moles
Moles CS2 required = 0.2/2 = 0.1
Mass CS2 = moles x molar mass = 0.1 x 76 = 7.6 g
% CS2 = (mass CS2/ Total Mass sample) x 100 = (7.6/13.82) x 100 = 55 %
b) PV = nRT V = nRT/P = (0.2 x 8.314 x (25+ 273))/90 = 5.5 L
2 marks for the balanced equation, 1mark for volume of liquid, 1 mark for volume
of SO2 produced