Relative Lowering of Vapour Pressure and Molar Mass Relation

Determination of molar mass of non-volatile solute from relative lowering of vapour pressure

Relative lowering of vapour pressure measured experimentally can be used to determine the molar mass of a non-volatile solute. A known mass of the non-volatile solute is dissolved in a known amount of solvent.

Let w gm of a solute of molecular weight m is dissolved in W gm of solvent of molecular weight M. Assume P_o be the vapour pressure of pure solvent at t°C and P be the vapour pressure of the solution at the same temperature.
Thus, the relative lowering of vapour pressure = [P_o − P] / P_o
Number of moles of solute dissolved (n) = w/m
Number of moles of solvent taken (N) = W/M
Total number of moles of solution = w/m + W/M
We know that mole fraction of solute = n/(n + N)
where, n = number of moles of solute
N = number of moles of solvent
Putting the value of n and N, we get-
Total number of moles of solution = w/m/[w/m + W/M]

According to Raoult's law, we know that the relative lowering of vapour pressure is equal to mole fraction of solute for a solution containing non-volatile solute.
Thus,
[P_o − P] / P_o = n/(n + N)
= w/m/[w/m + W/M]
In case of dilute solution,
n <<< N
so, [P_o − P] / P_o = wM/mW

Thus, from the above expression we can easily calculate the molecular weight of the solute. The determination of molecular mass by this method is often difficult because the accurate determination of lowering of vapour pressure is difficult.

The relative lowering of vapour pressure produced by dissolving 7.2 gm of a substance in 100 gm water is 0.00715. What is the molecular mass of the substance?

Answer

Hints:

Molecular mass of the substance (m) = [(7.2 × 18) / 0.00715] × 100 = 181.26 amu.