Question
What do you mean by elevation of boiling point and molal elevation constant? Give the formula or equation to determine the molar mass from elevation of boiling point.
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Answer
Boiling point of a liquid, is characteristic temperature at which its vapour pressure becomes equal to the atmospheric pressure. The boiling of a liquid depends on the atmospheric pressure. A liquid, therefore, boils at relatively lower temperature. Also liquid boils at relatively higher temperature at places where pressure is relatively higher. As the vapour pressure of a solution containing a non-volatile solute is lower than that of the solvent, its boiling point is higher than that of the solvent. The increase in boiling point is known as elevation in boiling point.
$ \Delta T_{b} = T_{1} - T_{0} $
Since for a solution, decrease in vapour pressure $ \propto $ m (molality) of solution. Thus, $ \Delta T_{b} \propto m $
and $ \Delta T_{b} = K_{b} \times m = \frac{K_{b} \times w_{2} \times 1000}{M_{2} \times w_{1}} $
To find molecular mass of solute
$ M_{2} = \frac{K_{b} \cdot w_{2} \cdot 1000}{\Delta T_{b} \cdot w_{1}} $
where $ K_{b} $ is molal elevation constant or ebullioscopic constant. $ K_{b} $ has units of K/m or K kg $ mol^{-1} $. For water $ K_{b} = 0.52~kg~mol^{-1} $.
$ \Delta T_{b} = T_{1} - T_{0} $
Since for a solution, decrease in vapour pressure $ \propto $ m (molality) of solution. Thus, $ \Delta T_{b} \propto m $
and $ \Delta T_{b} = K_{b} \times m = \frac{K_{b} \times w_{2} \times 1000}{M_{2} \times w_{1}} $
To find molecular mass of solute
$ M_{2} = \frac{K_{b} \cdot w_{2} \cdot 1000}{\Delta T_{b} \cdot w_{1}} $
where $ K_{b} $ is molal elevation constant or ebullioscopic constant. $ K_{b} $ has units of K/m or K kg $ mol^{-1} $. For water $ K_{b} = 0.52~kg~mol^{-1} $.
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