Question

Derive an expression to show that the relative lowering of vapour pressure is a colligative property. On the basis of this expression, how would you determine the molecular mass of a solute?

Answer 1

If p° represents the vapour pressure of a pure solvent and p represents the vapour pressure of the solution, we have

Lowering of vapour pressure = p° − p

Relative lowering of vapour pressure = `(p^circ - p)/p^circ`

The relative lowering of vapour pressure as well as the lowering of vapour pressure are colligative properties and depend upon the number of solute particles but not upon its nature.

The relative lowering of vapour pressure is equal to the mole fraction `chi_"solute"` of solute, i.e.,

`(p^circ - p)/p^circ = chi_"solute"`    ...(i)

From the above equation, it is clear that the relative lowering of vapour pressure of a solution depends only upon the mole fraction of the solute and is independent of its nature. Since mole fraction represents the number of moles of solute dissolved in a definite number of moles of the solvent, the relative lowering in vapour pressure is a colligative property.

If the solution contains n moles of solute dissolved in N moles of solvent, we have

`chi_"solute" = n/(N + n)`

Hence, the equation (i) can be written as

`(p^circ - p)/p^circ = n/(N + n)`    ...(ii)

Suppose w = mass of solute, W = mass of solvent, M = molecular mass of solvent, and M' = molecular mass of solute.

Therefore,

`n = w/(M')` and

`N = W/M`

Substituting this values in equation (ii), we have

`(p^circ - p)/p^circ = (w//M')/((W//M) + (w + M'))`    ...(iii)

Knowing the relative lowering of vapour pressure and the values of w, Wand M; the molecular mass M' of the solute can be calculated with the help of above equation.

For a very dilute solution, equation (iii) can be simplified as follows.

The amount of solute is significantly less than the amount of solvent in a very diluted solution. Therefore, n << N for an extremely diluted solution. Therefore, we might have

N + n ≈ N

In this situation, Eq. (ii) can be written as

`(p^circ - p)/p^circ = n/N`    ...(iv)

Substituting the values of n and N, we have

`(p^circ - p)/p^circ = (w//M')/(W//M)`

or `(p^circ - p)/p^circ = (wM)/(WM')`    ...(v)

which gives, `M' = (wM)/(W[(p^circ - p)/p^circ]`    ....(vi)

Knowing the values of mass of solute (w), mass of solvent (W), molecular mass of solvent (M) and relative lowering in vapour pressure `[(p^circ - p)/p^circ]`, the molecular mass M' of the solute can be calculated with the help of equation (vii), provided the solution is dilute.