Vapour Pressure Of A Solution

Relative lowering of vapour pressure of a solvent is a colligative property equal to the vapour pressure of the pure solvent minus the vapour pressure of the solution. For example, water at 20°C has a vapour pressure of 17.54 mmHg. Ethylene glycol is a liquid whose vapour pressure at 20°C is relatively low, an aqueous solution containing 0.010 mole fraction of ethylene glycol has a vapour pressure of 17.36 mmHg. Thus the vapour pressure lowering, $\Delta$P = 17.54 mmHg ¾ 17.36 mmHg = 0.18 mmHg.

RAOULT'S LAW

Vapour pressure of a number of binary solution of volatile liquids such as benzene and toluene at constant temperature gave the following generalization which is known as the Raoult's law.

The partial pressure of any volatile component of a solution at any temperature is equal to the vapour pressure of the pure component multiplied by the mole fraction of that component in the solution.

Suppose a binary solution contains n_A moles of a volatile liquid A and n_B moles of a volatile liquid B, if P_A and P_B are partial pressure of the two liquid components, the according to Raoult's law

$P_A=x_A.P_A^{0}and{P}_B=x_B.P_B^0$

Where x_A­ is the mole fraction of the component A given by n_A/n_A+ n_B; x_B is the mole fraction of the component B, given by n_B/n_A+n_B and $P_A^{0}and{P}_B^0$ are the vapour pressures of pure components A and B respectively.

If the vapour behaves like an ideal gas, then according to Dalton's law of partial pressures, the total pressure P is given by

$P=P_A+P_B=x_A^{}{P}_A^0+x_B^{}{P}_B^0$

= $\frac{n_A.P_A^0}{n_A+n_B}+\frac{n_B.P_B^0}{n_A+n_B}$

$\frac{n_A.P_A^0+n_B.P_B^0}{n_A+n_B}$

The relationship between vapour pressure and mole fraction of an ideal solution at constant temperature is shown. The dashed lines I and II represent the partial pressure of the components. (It can be seen from the plot that p_A and p_B are directly proportional to x_A and x_B respectively). The total vapour pressure is given by line marked III in the figure.

Illustration 1.

The vapour pressure of ethanol and methanol are 44.5 mm and 88.7 mm Hg respectively. An ideal solution is formed at the same temperature by mixing 60 g of ethanol with 40g of methanol. Calculate total vapour pressure of the solution.

Solution:

Number of moles of ethanol = $\frac{60}{40}$ = 1.5

Number of moles of methanol = $\frac{40}{32}$ = 1.25

X_A = $\frac{1.25}{1.25+1.3}$ = 0.4545 and X_B = 1–0.4545 = 0.545

Let A = CH₃OH, B = C₂H₅OH

Total pressure of the solution

P_T = X_A$P_A^o$ + X_B$P_B^o$

= 0.4545 ´ 88.7 + 0.545 ´ 44.5 = 40.31 + 24.27 = 64.58 mm 4g

Measurement of Vapour Pressure Lowering (Ostwald and Walker's Apparatus)

In this method a stream of dry air is bubbled successively through (i) the solution (ii) the pure solvent and (iii) a reagent which can absorb the vapour of the solvent. As the solvent is usually water the reagent is generally anhydrous Calcium Chloride. The complete assembly is shown in the figure.

The first three bulbs contain a weighed amount of the solution under examination and the next three bulbs contain a weighed amount of the pure solvent. A weighed amount of anhydrous calcium chloride is taken in the set of U-tubes at the end.

All the bulbs must be kept at the same temperature and air must be bubbled gradually to ensure that it gets saturated with the vapours in each bulb.

The dry air, as it passes through the solution, takes up an amount of vapour which is proportional to the vapour pressure of the solution at the prevailing temperature. This moist air passes through water (solvent), it takes up a further amount of vapour which is proportional to the difference in vapour pressure of the pure solvent and the solution.

It is evident that,

Loss in mass of solution $\propto P$

Loss in mass of solvent $\propto P-P$

Loss in mass of solution + loss in mass of solvent

$\propto P-P+P\propto P$

Where $P=V$ apour pressure of pure solvent, $P=V$apour pressure of solution

The calcium chloride tubes are weighed at the end of the experiment. The gain in mass should be equal to the total loss in mass of the solution and solvent which, in turn, is proportional to P⁰, as shown above.

In other words, $\frac{lossinmassofsolvent}{gaininmassofCaC{l}_{2}tubes}=\frac{P^{{}^{\circ }}-P}{P^{{}^{\circ }}}$

Thus, knowing the loss in mass of the solvent and gain in the mass of the calcium chloride tubes, it is possible to calculate the lowering of vapour pressure.

Illustration 2.

Dry air is passed through a solution containing 20g. of an organic non-volatile solute in 250 ml of water. Then the air was passed through pure water and then through a U-tube containing anhydrous CaCl₂. The mass lost in solution is 26g and the mass gained in the U-tube is 26.48 g. Calculate the molecular mass of the organic solute.

Solution:

Loss in mass in solution = 26g

Gain in mass in U-tube = 26.48 g.

Therefore, Loss in mass in solvent = 0.48g

We know that

$\frac{Lossinmassofsolvent}{gaininmassofU-tube}=\frac{p^0-p}{P^0}$

$\frac{0.48}{26.48}=\frac{P^0-P}{P^0}$= mole fraction of solute in the solution

$\frac{0.48}{26.48}=\frac{n_{solute}}{n_{solute}+n_{solvent}}=\frac{n_{solute}}{n_{solute}+250/18}$

$\frac{26.48}{0.48}=1+\frac{250}{18\times n_{solute}}$

$n_{solute}=\frac{250\times 0.48}{18(26)}$

$M_{solute}=\frac{20\times 18\times 26}{250\times 0.48}=78$

$\therefore$ Molecular mass of the organic solute = 78.