Chem1000A
Spring 2003
Assignment 9 - Solution
Two or three questions will be marked from each assignment.
DUE ON March 26, 2003 (Friday) 1:00 PM
To be dropped off at my office (C886)
Use SI units throughout your calculations. R = 8.3145 J K-1 mol-1
1. Specify the interparticular forces present in the following compounds:
(a) NH3
This molecule has a trigonal pyramidal molecular geometry and is polar, i.e., it has a dipole moment.
Intermolecular forces:
Dipole-dipole interactions, more specifically: hydrogen-bonding interactions
Dipole-induced dipole interactions
London dispersion forces
(b) SO2
This molecule has a bent molecular geometry and is polar, i.e., it has a dipole moment.
Intermolecular forces:
Dipole-dipole interactions
Dipole-induced dipole interactions
London dispersion forces
(c) Na2SO4
This compounds consists of Na+ cations and SO42- anions. The sulfate anion is tetrahedral and has no
dipole moment.
Intermolecular forces:
Ion-Ion interactions
Ion-induced dipole interactions
London dispersion forces
(d) CF4
This molecule has a tetrahedral molecular geometry and is non-polar. All the bond dipole moments
from the C-F bonds cancel out.
Intermolecular forces:
London dispersion forces
2. What interparticular forces are present in a solution of NaCl in H2O?
You have to consider three particles in this solutions: Na+, Cl-, and H2O
You have the following interparticular interactions:
Ion-ion between Na+ and ClIon-dipole interactions between Na+ and H2O, and Cl- and H2O.
Ion-induced dipole interactions between one ion and any other particle
Dipole-dipole between two water molecules
Dipole-induced dipole between water and any other particle
London dispersion forces between any two particles
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3. Which of the following compounds do you expect to be soluble in water, which ones in hexane?
Explain your answer briefly.
(a) Xe
Xenon is non-polar; you would expect it to be soluble in a non-polar solvent: hexane.
(b) Na2SO4
Solium sulfate is an ionic compound; you expect it to be soluble in a polar solvent: water.
(c) diethyl ether (CH3CH2OCH2CH3)
Diethyl ether has a small dipole moment, because of the bent arrangement about the oxygen, but has
non-polar groups attached to the oxygen. You would expect it to be soluble in non-polar solvents, like
hexane. It is also soluble in water because of the dipole moment.
(d) octane (CH3CH2CH2CH2CH2CH2CH3)
Octane is a non-polar molecule. It is soluble in non-polar solvents, such as hexane.
(e) SF6
Sulfur hexafluoride is a non-polar compounds; all S-F bond dipole moments cancel out. You would
expect that it is soluble in non-polar solvents, such as hexane.
4. Comparing the intermolecular forces present in pure SF4 and in pure TeF4. Which of these two
compounds has a higher vapour pressure at room temperature and which one has a higher boiling
point? One of the compounds is a gas and the other is a solid. Identify the solid among the two
tetrafluorides. Provide explanations for your answers.
Both tetrafluorides are polar compounds. Their molecular geometry is a seesaw.
The intermolecular forces present are
Dipole-dipole interactions
Dipole-induced dipole interactions
London dispersion forces
The dipole moment in TeF4 is larger than that in SF4, since the Te-F bond is more polar than the S-F
bond. Therefore, the dipole-dipole interactions are stronger in TeF4. TeF4 is bigger/ has more
electrons and, therefore, has a larger polarizability. With a larger polarizability, it has stronger
London dispersion forces. The dipole-ind. Dipole interactions are also stronger in TeF4.
The attractions between TeF4 molecules are much stronger than in SF4. As a consequence, TeF4 has a
lower vapour pressure and a higher boiling point than SF4.
The difference in strengths of the intermolecular forces results in different states of matter for these
two compounds. Tellurium tetrafluoride is a solid and sulfur tetrafluoride is a gas.
5. You place a bottle containing 1.00 L of water at 25.0 °C in a non-ventilated room, which has a
volume of 5.00 m × 5.00 m × 2.50 m. Then you open the bottle and wait long enough for the
establishment of equilibrium between the liquid and gas phase of water. Assume the air inside the
room is completely dry before you open the bottle.
(a) How much water will be left and what will be the partial pressure of water inside the room.
(b) If you have 2.00 L of water, instead of 1L, would the partial pressure of water change?
The vapour pressure of water at 25.0 °C is 0.03126 atm. Assume a density of 1 g/mL for water.
(a) You have to find out the number of moles that you have in the 1.00 L of water and the number of
moles that you have in the gas phase if you have the equilibrium vapour pressure of water. Then you
have to decide if the amount of liquid water is sufficient to fill the room with water up to its vapour
pressure or if you have a deficiency of water and all the water will go into the gas phase and will not
reach the equilibrium vapour pressure.
V = 62.5 m3, T = (25.0 + 273.15) K = 298.2 K, M(H2O) = 18.0153 g mol-1
p = 0.03126 atm × (101325 Pa/1atm ) =3167 Pa
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Assuming a density for water of 1 g/mL: m(H2O) = 1000 g
Inside the bottle: n(H2O) = 1000 g/(18.0153 g mol-1) = 55.5 mol
In the gas phase, saturated with water: pV = nRT,
n = pV/RT = 3167 Pa×62.5 m3/(8.3145 J K-1 mol-1 × 298.2 K) = 79.8 Pa m3 mol J-1
= 79.8 (kg m-1 s-2) m3 mol (kg-1 m-2 s2) = 79.8 mol
The number of moles in the gas phase, saturated with water would be much larger than the number
of moles of water available in the bottle. Therefore, all the water will evaporate and will not reach
the equilibrium vapour pressure of water at 25 °C.
Partial pressure of water with all the water from the bottle in the gas phase:
p = nRT/V = 55.5 mol × 8.3145 J K-1 mol-1 × 298.2 K/62.5 m3 = 2200 J m-3 = 2200 (kg m2 s-2) m-3
= 2200 Pa
(b) In the case that you have 2L of water, you double the moles of the water in the liquid state:
111 mol
Now, you have enough, even more than enough, water in the liquid state to reach the equilibrium
vapour pressure of water. The partial pressure of water will be different from (a)! The partial
pressure of water in the room will be 0.03126 atm =3167 Pa.
6. At a certain day in Lethbridge, water boils at 96.5 °C.
(a) How much NaCl do you have to add to exactly 1 kg of water to raise the boiling point of water in
the solution to 100.00 °C?
∆T =100.00 °C – 96.5 °C = 3.5 °C = 3.5 K (the difference is the same in K and °C)
∆T = Kb.p. ×ν ×molality , ν = 2 for NaCl
Molality = ∆T/ 2Kb.p. = 3.5 K/(2 × 0.512 K kg mol-1) = 3.4 mol/kg= mol of solute/mass of solvent
3.4 mol/kg × 1kg = 3.4 mol
M(NaCl) = 58.4425 g/mol
m(NaCl) = 58.4425 g/mol× 3.4 mol = 200 g of NaCl
(b) What is the vapour pressure of water above this solution at 25.00 °C?
M(NaCl) = 58.4425 g/mol , M(H2O) = 18.0153 g mol-1
n(NaCl) = 3.4 mol
n(H2O) = 1000 g/( 18.0153 g mol-1) = 55.5084 mol
Since NaCl dissociates in water into two particles:
n(solute particles) = 2 ×3.4 mol = 6.8 mol
X(H2O) = 55.5084 mol/62 mol = 0.89
p(H2O) = 0.89 × 0.03126 atm = 0.028 atm
The ebullioscopic and cryoscopic constants of water (in SI units) are 0.512 and 1.86 K kg mol-1,
respectively. The vapour pressure of water at 25.0 °C is 0.03126 atm.
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