11. Colligative Properties questions

Michael Mombourquette

11. Colligative Properties questions

What is the vapour pressure lowering of a solution containing 10.0g of sucrose (C₁₂H₂₂O₁₁) in 100g of water at 25ºC; the vapour pressure of water is 3.17 kPa.
The measured vapour pressure of a solution of 0.50 mol of potassium chloride (KCl) dissolved in 1.000 kg of water is lowered by 1.63 kPa at 100ºC, relative to pure water, which has a vapour pressure of 101.325 kPa at this temperature. Calculate a generic concentration of solutes from the vapour pressure lowering and compare that to the concentration of the solution you made here. (don’t use van’t Hoff). Why is van’t Hoff’s correction necessary?
A solution contains 0.025 mol of naphthalene in 25 g of benzene. At 25ºC, benzene has a vapour pressure of 12.7 kPa and naphthalene has a vapour pressure of 0.011 kPa, so we can approximate it to be non-volatile. Calculate the vapour pressure lowering, the boiling point elevation and the freezing point depression of the solution.
A solution contains 7.25 g of an unknown non-volatile solute in 100g of benzene. the boiling point of the solutio is 0.76ºC higher than that of pure benzene. What is the molar mass of the unknown compound.
What is the osmotic pressure of a 0.10M solution of sucrose in water at 20ºC?
1.50 g of a polymer is dissolved in enough water to make 85 mL of solution. The osmotic pressure is 375 kPa. what is the average molar mass of the polymer?

answers

The vapour pressure lowering of an aqueous solution of a non-volatile solute is calculated using the equation
$P_{solution} = \chi_{water}\times P^*_{water}$

The vapour pressure lowering will simply be $P*_{water} - P_{solution}$
We need to calculate the mole fraction of the water.

$n(water) = \frac{100.0\textrm{ g}}{18.015\textrm{g/mol}} = 5.551 \textrm{mol}$

$n(sucrose) = 10.0 \textrm{ g}}{342.30\textrm{g/mol}} = 0.02921 \textrm{mol}$

$\chi_{water} = \frac{5.551}{5.551+0.02921} = 0.99476$

$\[ P_{solution} = 0.99476\times 3.17 \textrm{ kPa} = 3.1534 \textrm{ kPa}$

$\Delta P^*_{solution} = 3.17 - 3.1534 = 0.016 \textrm{ kPa} = 16 \textrm{ Pa}$

As you can see, were pushing the sig figs a bit here. so clearly, the accuracy of the vapour pressure lowering calculation is not great.
We can use ratios of pressures as ratios of moles and so:
$\chi_{solute} = \frac{\Delta P^*_{solution}}{P^*_{water}} \frac{1.63}{101.325}=1.61\times10^{-2}$

A mass of 1.000 kg of water contains (1000/18.015)=55.51 mol water. The mole fraction of KCl is

$\chi_\textrm{KCl}=\frac{0.50}{0.50+55.51} = 8.9\times10^{-3}.$

The mole fraction of solutes is quite a bit larger than the mole fraction of KCl ( $\frac{1.61\times 10^{-2}}{8.9\times10^{-3}}=1.8$ )
This is due to the fact that KCl breaks up into its constituent ions. The rudimentary van’t Hoff Factor $i$ would be 2 but this data says the experimental ratio is 1.8, more likely do to other complicating factors.
let’s calculate the mole fraction of the naphthalene. the Molar mass of benzene is 78.113 g/mol, so the moles of benzene is
$\frac{85.0}{78.113\textrm{ g/mol}} = 1.088\textrm{ mol}$

so the mole fraction is

$\chi_{naph}=\frac{.025}{.025+1.088}=0.022$

So, the vapour pressure lowering is

$\DeltaP^* = 0.022\times 12.7 = 0.28\textrm{ kPa}$

the molality of naphthalene is

$m_{naph}= \frac{0.025\textrm{ mol naph}}{0.850 \textrm{ kg benzene}}=0.294 \textrm{ mol/kg}.$

The boiling point elevation using the molal boiling point elevation constant from Table 1 in the chapter.

$\Delta T_b = 2.53\mathrm{^\circ C\;mol^{-1}\;kg}\times 0.294 \mathrm{mol\;kg^{-1}}=0.74\mathrm{^\circ C}$

the freezing point depression can be calculated similarly

$\Delta T_f = 4.90\mathrm{^\circ C\;mol^{-1}\;kg}\times 0.294 \mathrm{mol\;kg^{-1}}=1.44\mathrm{^\circ C}$
The boiling point elevation constant for benzene, from the table, is $2.53\mathrm{^\circ C\;mol^{-1}}$ and the boiling point elevation is $0.76 \mathrm{^\circ C\;mol^{-1}\;kg}$ , so the molality of the solution is
$m=\frac{\Delta T_b}{K_b}= \frac{0.76 \mathrm{^\circ C}}{2.53 \mathrm{^\circ C\;mol^{-1}\;kg}} = 0.30 \mathrm{\;mol^{-1}\;kg}}$

since the solvent total mass is only .1 kg, the amount of solute actually present will be only $0.030 \mathrm{\;mol^{-1}\;kg}}$ . So, the molar mass is $\frac{7.25\textrm{ g}}{0.030\textrm{ mol}}=242\textrm{g/mol}$ .
$1.00 \textrm{ L} = 1.00\times10^{-3}\mathrm{\;m^3}$ . this volume of solution contains 0.10 mol of solute. So:
$\pi = \frac{0.10\textrm{ mol}\times8.314\mathrm{\;J\;K\;mol^{-1}}\times298\textrm{ K}}{1.0\times10^{-3}\mathrm{\;m^3}}=244\textrm{ kPa}$

This is much higher than atmospheric pressure of about 100 kPa.
the osmotic pressure is 375 Pa and the volume is $85.0\times10^{-6}\mathrm{\;m^3}$ . we can rewrite the equation for osmotic pressure to get the number of moles of solute:
$n=\frac{375\textrm{ Pa}\times85.0\times10^{-6}\mathrm{\;m^3}}{ 8.314\mathrm{\;J\;K^{-1}\;mol^{-1}}\times289\textrm{ K}}=1.29\times10^{-5}\textrm{ mol}$

So, the molar mass of the compound is

$M=\frac{1.50 \textrm{ g}}{1.29\times10^{-5}\textrm{ mol}}=1.16\times10^5\textrm{ g/mol}$