Long Answer TypeHeptane and Octane form an ideal solution at 373 K. The vapour pressures of the pure liquids at this temperature are 105.2 K Pa and 46.8 K Pa respectively. If the solution contains 25 g of heptane and 28.5 g of octane, calculate
(i) Vapour pressure exerted by heptane.
(ii) Vapour pressure exerted by solution.
(iii) Mole fraction of octane in the vapour phase.
Short Answer TypeLong Answer TypeEvery pure liquid exerts a vapour pressure in the space above it. This is the vapour pressure of the solvent over it at that particular temperature. It depends upon the nature of the solvent and the temperature.
(a) If a volatile solute is dissolved, vapour pressure of the solvent is increased.
(b) However, if a non-volatile solute is dissolved in it, the vapour pressure of the solution is lowered. This is because, in a solution, the percentage of the volatile solvent molecules, which only contributes towards vapour pressure is diminished.
Fig. Decrease of vapour pressure when a non-volatile solute is added to the solvent.
Since, the solute molecules are non-volatile and show no measurable tendency to escape from the solution as vapour, consequently, the vapour pressure of a solution is always lower than that of its solvent.
Raoult’s gave a relation between the relative lowering of vapour pressure and the mole fraction of the solute. Mathematically:
(mole fraction of the solute)
Using the above equation, we can determine the molecular weight of the solute, when the lowering in v.p. is known, when a known weight of the solute w, dissolved in a known wt. of the solvent W.p0 is the vapour pressure of the pure solvent and m and M are the molecular weights of solute and solvent respectively.
Short Answer Type
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