Question

Vapour-pressure of pure benzene at any temperature is 640 mm Hg. A non-volatile, non-electrolyte solid whose weight is 2.175 g, is added to 39.0 g of benzene. Vapour pressure of solution is 600 mm Hg. Calculate the molecular weight of the solid.

Answer 1

To solve this question first we have to understand the term colligative properties. When a non-volatile solute is added to a pure solvent, its vapour pressure decreases. Relate this by vapour pressure and number of moles to calculate the weight of the compound.

Complete step by step answer:
Given in the question, pure benzene is the pure solvent to which a non-volatile solute is added.
- Now, we know that:
Raoult’s law states that the relative lowering of vapour pressure of a solution containing a non-volatile solute is equal to the mole fraction of the solute in the solvent.
Writing mathematical form of Raoult’s law –
$\dfrac{{{p}_{\circ }}-{{p}_{s}}}{{{p}_{\circ }}}=\dfrac{n}{n+N}$
Where,
${{p}_{s}}$ = vapour pressure of solution= 600 mm Hg (given)
${{p}_{0}}$ = vapour pressure of pure solvent = 640 mm Hg (given)
n = number of moles of the solute (unknown)
N = number of moles of the solvent (benzene)
- We can calculate the number of moles by the ratio of given mass to the molar mass.
Given, weight of the Unknown compound = 2.175g
- Let’s assume the molar mass of the compound be m
$n=\dfrac{2.175}{m}$
Given weight of benzene- 39 g (given)
We know the molar mass of benzene = 78 g
$n=\dfrac{39}{78}=0.5$
Substituting these values in the formula, we get –

& \dfrac{640-600}{640}=\dfrac{\frac{2.175}{m}}{\dfrac{2.175}{m}+0.5} \\\ & \dfrac{640-600}{640}=\dfrac{2.175}{2.175+0.5m} \\\ \end{aligned}$$ m = 65.25 Therefore, the answer is – The molecular weight of the solid is 65.25 g. **Note:** Relative lowering of vapour pressure is defined as the ratio of lowering of vapour pressure to the vapour pressure of the pure solvent. It can be expressed in terms of Raoult’s law.