PRACTICE EXAM B ANSWERS – 11/28/06
1. Imagine a jar with an ideal gas in it. The volume is 20 L, the number of moles is 2.0, the pressure is 1 atm and the temperature is 5 K. a. If the number of moles is changed to 4.0, the volume would be doubled (40 L).
b. If the pressure is changed to 0.5 atm, the volume would be doubled (40 L).
c. If the temperature is changed to 1 K, the volume would be lessened by a factor of 5 (4
L).
1. At STP, when 5.40 mol HCl reacts in the following equation, how many liters of H
2 would you end up with?
Zn
(s)
+ 2HCl
(aq)
→ ZnCl
2 (aq)
+ H
2 (g)
5.40 mol HCl x 1 mol H
2
x 22.4 L = 60.5 L H
2
2 mol HCl 1 mol
3. How many liters of H
2
would be formed at 742 mm Hg and 15ºC if 25.5 g of zinc was allowed to react in the above equation?
PV = nRT V = (nRT)/P
P = 742 mm Hg x 1 atm = 0.976 atm
760 mm Hg n = 25.5 g Zn x 1 mol Zn x 1 mol H
2
= 0.390 mol H
2
65.39 g Zn 1 mol Zn
(It MUST be in mol H
2
because you are looking for Liters of H
2
, not Liters of Zn!!!)
R = 0.0821 L x atm /
K x mol
T = 15ºC + 273 = 288 K
V = (0.390 mol H
2
x 0.0821
L x atm
/
K x mol
x 288 K) / (0.976 atm) = 9.45 L H
2
3. What is the molar mass of the gas with a density of 1.342 g/L at STP?
1.342 g x 22.4 L = 30.1 g/mol
1 L 1 mol
5. An unknown gas is placed in a 1.500 L bulb at a pressure of 356 mmHg and a temperature of 22.5ºC, and is found to weigh 0.9847 g. What is the molecular mass of the gas? n = PV/RT = [(356 mmHg x 1.00 atm/760 mmHg)(1.500 L)] / [(0.0821)(273 + 22.5ºC)] n = 0.0290 mol molar mass = 0.9847 g / 0.0290 mol = 34.0 g/mol
6. Determine what the forces of attraction are in the following molecules: a. CHF
3 b. O
2
London dispersion, dipole-dipole
London dispersion c. C
12
H
26
London dispersion d. CH
3
OH London dispersion, dipole-dipole, hydrogen-bonding e. HF London dispersion, dipole-dipole, hydrogen-bonding
7. Of the two substances O
2
and C
12
H
26
, which one has higher vapor pressure at the same temperature? Why?

O
2
because it has weaker intermolecular forces, thereby making it easier for O
2
molecules to escape the liquid to become gas (vapor), thus increasing the vapor pressure. We know the smaller substance of the two has weaker forces because both substances have only
London dispersion forces, which is dependent on size (smaller means less points of contact which means weaker forces). O
2
is the smallest and thus has weaker forces.
8. Of the two substances HF and C
12
H
26
, which one has a lower boiling point? Why?
C
12
H
26
because it has weaker intermolecular forces and the individual molecules therefore have an easier time of leaving the liquid to become gas, so boiling can happen at a lower temperature. We know C
12
H
26
has weaker forces because it has only London dispersion forces, while HF has dipole-dipole and hydrogen-bonding forces as well as
London dispersion forces.
9. Of the two substances CHF
3
and O
2
, which one has a higher viscosity and surface tension? Why?
CHF
3
because it has stronger intermolecular forces, so the molecules stay together more and resist flowing or expanding the surface of its liquid. We know CHF
3
has stronger forces because O
2
has only London dispersion forces, while CHF
3
has dipole-dipole as well as London dispersion forces.