Question

A solution containing 8 g of a substance in 100g of diethyl ether boils at ${36.86^ \circ }$ C. Whereas pure ether boils at ${35.60^ \circ }$ C. Determine the molecular mass of the solute. (For ether, ${K_b}$ = 0.02k kh ${\text{mo}}{{\text{l}}^{ - 1}}$).

Answer 1

Colligative properties are the properties of solution. Colligative properties depend only on the number of solute particles and not on the nature of solute particles. But it depends on the nature of solvent particles.

Formula Used:
$\Delta {T_b} = i{K_b} \times m$
where $\Delta {T_b}$ is the difference in the temperature between solution and pure solvent. 'i' is the van’t Hoff factor.

Complete step by step answer:
In the given question, the colligative property to be used is elevation in boiling point. Elevation in boiling point is a phenomenon in which the boiling point of a solvent will be higher when compound is added to it as compared to the boiling point of pure solvent. This happens when non-volatile solute is added to pure solvent. The formula to be used here is,
$\Delta {T_b} = i{K_b} \times m$
where $\Delta {T_b}$ is the difference in the temperature between solution and pure solvent. 'i' is the van’t Hoff factor. It accounts for the number of ions formed by the compound in the solution. For non-electrolyte the value of i is 1. Here the value of i is 1. ${K_b}$ is the ebullioscopic constant or boiling point elevation constant. 'm' is the molality. Molality of a solute is defined as number of moles of solute divided by weight of solvent in kg.
$\Delta {T_b} = 36.86 - 35.60 = {1.26^ \circ }$ C
i = 1
${K_b}$ = 2.02
m = $\dfrac{{{\text{Number of moles}}}}{{{\text{Weight of solvent }}\left( {{\text{in kg}}} \right)}}$
where the weight of solvent is $\dfrac{{100}}{{1000}}$ kg.
Hence, m = $\dfrac{{{\text{Number of moles}}}}{{\dfrac{{100}}{{1000}}}} = 10 \times {\text{Number of moles = 10}} \times \dfrac{{{\text{Given weight of solute}}}}{{{\text{Molar mass of solute}}}} = \dfrac{{10 \times 8}}{{{\text{Molar mass of solute}}}}$
$\Delta {T_b} = i{K_b} \times m$
Therefore, $1.26 = 1 \times 2.02 \times \dfrac{{10 \times 8}}{{{\text{Molar mass of solute}}}}$
Molar mass of solute = $\dfrac{{80 \times 2.02}}{{1.26}} = 128.25$ g
Hence the molar mass of solute is $128.25$ g.
Note:
The boiling point can be measured accurately using an ebullioscope.
There are four colligative properties of solution which are as follows,
-Elevation in boiling point
-Depression in freezing point
-Osmotic pressure
-Relative lowering of vapour pressure