What do you understand by elevation of boiling point and molal elevation constant, with the help of this constant how can you find the molecular mass of non volatile solute?

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Solution

Elevation of Boiling Point: Elevation of boiling point on adding one mole solute to $1000$ $g$ solvent is called molal boiling point elevation constant. It is denoted by $K_{b}$ . Its unit is $K$ $k g$ ${m o l}^{- 1}$ , e.g. for water $K_{b} = 0.52$ $K$ $k g$ ${m o l}^{- 1}$ .
Derivation of relation: Increase in boiling point of the solvent, on adding non-volatile solute to the solvent is called elevation of boiling point.
$Δ T_{b} = T_{3} - T_{2}$
where $T_{2} =$ boiling point of solvent, and $T_{3} =$ boiling point of solution.
Experimentally, it is observed that elevation in boiling point is directly proportional to molal concentration of solute dissolved in definite amount of the solvent.
$Δ T_{b} \propto m$
$Δ T_{b} = K_{b} \times m$ .....(1)
$K_{b}$ is proportionality constant called molal boiling point elevation constant. We know that molality of a solution is number of moles of solute in $1000 g$ solvent. It is denoted by $m$ .
Determination of Molecular mass of a non-volatile solute:
Say molecular mass of the solute is $M_{1}$ . $W_{1} g$ solute is present in $W_{2} g$ solvent. Molality of solution is
$m = \frac{W_{1}}{M_{1}} \times \frac{1000}{W_{2}}$ .........(2)
On putting value of $m$ from equation (2) to (1),
$Δ T_{b} = \frac{K_{b} \times W_{1} \times 1000}{M_{1} \times W_{2}}$
$M_{1} = \frac{K_{b} \times W_{1} \times 1000}{Δ T_{b} \times W_{2}}$
Molecular mass of the solute may be determined by this formula.

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