School of Chemistry and Physics, University of KwaZulu-Natal, Westville Campus, Durban
CHEMICAL REACTIVITY – CHEM120
Tutorial 3 – 15 and 17 August 2012
1. (a) Determine the mole fraction and molality of a solution made by dissolving 40.0 g
of ammonium phosphate (NH4)3PO4 in 200.0 g of water?
Moles of ammonium phosphate = 40.0 g/149.12 g mol-1 = 0.268 mol
Moles of water = 200.0 g/18.02 g mol-1 = 11.10 mol
Mole fraction: xammonium phospate = 0.268 mol/(0.268 + 11.10) mol = 0.024
Molality:
m = 0.268 mol/0.2 kg = 1.34 m
(b) Calculate the vapour pressure (in mmHg) of the water over the solution at 90°C
(vapour pressure of water at 90°C is 525.8 mmHg)
xwater = 11.10 mol/(0.268 + 11.10) mol = 0.976
Pwater = xwater × Powater = 0.976 × 525.8 mmHg = 513.18 mmHg
2. The density of a 2.45 M aqueous solution of methanol (CH3OH) is 0.976 g/mL. What
is the molality of the solution? The molar mass of methanol is 32.04g mol-1.
We assume 1 L of solution, so the number of moles of methanol would be 2.45 mol.
Mass of water in 1 L of soln. = density x vol. = 0.976 × 1000 mL = 976 g
Mass of water = mass of soln – mass of solute = 976 g – (2.45 mol × 32.04 g mol-1) = 898 g
Molality of soln = 2.45 mol of CH3OH/ 0.898 kg H2O = 2.73 m
3. What is the boiling point of a solution containing 0.154 g of camphor (C10H16O)
dissolved in 11.3 g of benzene? (Kbp = 2.53°C m-1 and Boiling point of benzene =
80.1°C)
Moles of camphor = 0.154 g/152.3 g mol-1 = 1.01 × 10-3 mol
Molality = 1.01 × 10-3 mol/1.13 × 10-2 kg = 8.94 × 10-2 m
Tbp = Kbp × molality  Tbp = 2.53°C m-1 × 0.089 m = 0.225°C
Boiling point of solution = 80.1°C + 0.225°C = 80.325°C
4. What mass of ethylene glycol, CH2(OH)CH2(OH) must be added to 2.505 kg of water
to lower the freezing point of the water from 0°C to -7.9°C? (Kfp = 1.86°C m-1)
From Tfp = Kfp × molality
molality = Tfp/ Kfp  m = 7.9°C/1.86°C m-1 = 4.2 m
molality = nethylene glycol/mwater
nethylene glycol = 4.2 mol kg-1 × 2.505 kg = 10.5 mol
methylene glycol = 10.5 mol × 62.1 g mol-1 = 652 g = 0.652 kg
5. Consider the following phase diagram of water:
(a) At what temperature and pressure are the three phases of water in equilibrium?
0.00980C and 0.00603 atm
What is this point called? Triple point of water
(b) What is the normal melting point and boiling point of water at a pressure of 1 atm?
00C and 1000C
(c) At a pressure of 218 atm what would be the effect of a decrease in temperature on
the gaseous phase of water?
The gaseous phase of water would move towards the liquid phase