Total Pressure of a Solution will be:
Raoult's Law as a special case of Henry's law
Example 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.
Number of moles of ethanol = 60/40 = 1.5
Number of moles of methanol = 40/32 = 1.25
XA = 1.25/1.25 + 1.3 = 0.4545 and XB = 1 – 0.4545 = 0.545
Let A = CH3OH, B = C2H5OH
Total pressure of the solution
PT = XAP0A + XBP0B
= 0.4545 × 88.7 + 0.545 × 44.5 = 40.31 + 24.27
= 64.58 mm Hg
Example 2. The composition of vapour over a binary ideal solution is determined by the composition of the liquid. If XA and YA are the mole-fraction of A in the liquid and vapour, respectively find the value of XA for which YA − XA has a minimum. What is the value of the pressure at this composition.
Subtracting xA from both the sides, we get
Now differentiating w.r.t. xA, we get
The value of xA at which yA − xA has a minimum value can be obtained by putting the above derivative equal to zero. Thus we have
Solving for xA, we get
Example 3. A very small amount of non-volatile solute (that does not dissociate) is dissolved in 56.8 cm3 of benzene (density 0.889 g cm-3). At room temperature, the vapour pressure of this solution is 99.88 mm Hg while that of benzene is 100mm Hg. Find the molality of this solution. If the freezing temperature of this solution is 0.73 degree lower than that of benzene. What is the value of molal freezing point depression constant of benzene?
ΔTf = Kf × molality f
∴ 0.73 = K × 0.1452
fK = 5.028 K. molality
Example 4. What is the composition of the vapour which is in equilibrium at 30°C with a benzene-toluene solution with a mole fraction of benzene of 0.400?
(PoB = 119 torr and PTo = 37.0 torr)
Total pressure of the solution is given by
PT = XB P0B + XTP0T
= 0.4 × 119 + 0.6 × 37
= 47.6 + 22.2
= 69.8 torr
Applying Dalton’s law for mole fraction in vapour phase.
YB = PB/PT = 47.6/62.4
YT = 1– 0.763
Q.1. If PA is the vapour pressure of a pure liquid A and the mole fraction of A in the mixture of two liquids A and B is x, then, what is the partial vapour pressure of A?
Q.2. Vapour pressure of C6H6 and C7H8 mixture at 50°C is given by P (mm Hg) = 180 XB + 90, where XB is the mole fraction of C6H6. A solution is prepared by mixing 936 g benzene and 736 g toluene and if the vapours over this solution are removed and condensed into liquid and again brought to the temperature of 50°C, what would be mole fraction of C6H6 in the vapour state?
Q.3.When a solute is added to a pure solvent, which statement is not true?
(a) Vapour pressure of the solution becomes lower than the vapour pressure of the pure solvent.
(b) Rate of evaporation of the pure solvent is reduced.
(c) Solute does not affect the rate of condensation.
(d) None of these.
Q.4. According to Raoult's law the relative decrease in solvent vapour pressure over the solution is equal to
(a) The mole fraction of the solvent.
(b) The number of moles of the solute.
(c) 'i' times the mole fraction of the solute which undergoes dissociation or association in the solvent (i = Vant Hoff factor).