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Difference between Internal Energy and Enthalpy

A dilute solu...

Question

A dilute solution Contains “x” moles of solute “A” in 1 kg of solvent with molal elevation constant kb. The solute dimerises in the solution according to the following equation. The degree of association is $(α)$
$2 A ⇌ A_{2}$

The degree of association is equal to

$α = \frac{(K_{b} . x - Δ T_{b})}{Δ T_{b} .2}$

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$α = \frac{2 (K_{b} . x - Δ T_{b})}{K_{b} . x}$

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$α = \frac{2. Δ T_{b}}{K_{b} . x} + 2$

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$α = \frac{Δ T_{b}}{2. K_{b} . x}$

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Solution

The correct option is B
$α = \frac{2 (K_{b} . x - Δ T_{b})}{K_{b} . x}$

A dilute solution Contains “x” moles of solute “A” in 1 kg of solvent with molal elevation constant kb. The solute dimerises in the solution according to the following equation. The degree of association is $(α)$
$2 A ⇌ A_{2}$

$Δ T_{b} = i \times K_{b} \times \frac{x}{1} \frac{Δ T_{b}}{K_{b} . x} = 1 - \frac{\propto}{2} \frac{\propto}{2} = 1 - \frac{Δ T_{b}}{K_{b} x} \propto= 1 (1 - \frac{Δ T_{b}}{K_{b} . x}) \propto= 2 (\frac{K_{b} . x - Δ T_{b}}{K_{b} . x})$

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