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Elevation in Boiling Point - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:35 AM | #JEE Main

Quick Facts

  • Elevation in Boiling Point is considered one the most difficult concept.

  • 14 Questions around this concept.

Concepts Covered - 1

Elevation in Boiling Point

Elevation in Boiling Point:

Boiling Point: It is the temperature of a liquid at which its vapour pressure becomes equal to the atmospheric pressure. 

Now, the lowering of the vapour pressure of the solution occurs when the addition of non-volatile solute in solvent happens. In order to boil the solution, it is necessary to increase the temperature of the solution above the boiling point of the pure solvent. It means the boiling point of the solution is always higher than the boiling point of the pure solvent. This increase in the boiling point of the solution is called elevation in boiling point\left ( \Delta T_b \right )

  • It is also termed as Ebullioscopy.

  • Suppose \mathrm{T^o_b} and \mathrm{T_b} are the B.P. of pure solvent and solution respectively, then elevation in B.P \left (\Delta \mathrm{T_b} \right ) is given as \Delta \mathrm{T_b} = \mathrm{T_b} - \mathrm{T_b^o}

 

\Delta \mathrm{T_b} is directly proportional to the molality of solution

\\\mathrm{\Delta T_b\: \propto \: m} \\ \mathrm{\Delta T_b\:=K_bm}

  • If molality of the solution is one, then

\Delta \mathrm{T}_{\mathrm{b}}=\mathrm{K}_{\mathrm{b}}

  • The elevation in B.P. is also given as
    \Delta\mathrm{T}_{\mathrm{b}}=\mathrm{K}_{\mathrm{b}}\times\frac{\mathrm{w}}{\mathrm{M}} \times \frac{1000}{\mathrm{W}}

  • Molecular weight of solute can be found out as follows
    \mathrm{M=\frac{K_{b} \times w \times 1000}{\Delta T_{b} \times W}}

Here w = weight of solute

 

W = weight of solvent
Kb= molal elevation constant or ebullioscopic constant.
M = molar mass of solute

M1 = molar mass of solvent

The value of K_b or Ebullioscopic constant is a property of the solvent only and does not depend on the type of solute. The value of K_b can be calculated as: 

\mathrm{K}_{\mathrm{b}}=\frac{\mathrm{RM_{1}} \mathrm{T}^{2}}{1000 \: \mathrm{Lv} \: \mathrm{or}\: \Delta \mathrm{Hv}}

Here Lv or \Delta\mathrm{T}_{\mathrm{b}}=\mathrm{K}_{\mathrm{b}}\times\frac{\mathrm{w}}{\mathrm{M}} \times \frac{1000}{\mathrm{W}}= latent heat of vaporization.

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Elevation in Boiling Point

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