(C) The elevation in boiling point is given by $\Delta T_{b} = T_{b} - T_{b}^{0} = 100.52^{\circ} C - 100^{\circ} C = 0.52^{\circ} C$.For dimerization

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(C) The elevation in boiling point is given by $\Delta T_{b} = T_{b} - T_{b}^{0} = 100.52^{\circ} C - 100^{\circ} C = 0.52^{\circ} C$.For dimerization,the van't Hoff factor is $i = 1 - \alpha + \frac{\alpha}{2} = 1 - \frac{\alpha}{2}$.Using the formula $\Delta T_{b} = i \times K_{b} \times m$,where $m = 2 \, M$:$0.52 = (1 - \frac{\alpha}{2}) \times 0.52 \times 2$.Dividing both sides by $0.52$,we get $1 = (1 - \frac{\alpha}{2}) \times 2$.$1 = 2 - \alpha$.$\alpha = 1$.Thus,the percentage association is $1 \times 100 = 100 \, \%$.

Class 12 Chemistry Solutions Abnormal molecular mass

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$A$ solute $A$ dimerizes in water. The boiling point of a $2 \, M$ solution of $A$ is $100.52^{\circ} C$. The percentage association of $A$ is ..... $\%$.
(Round off to the Nearest integer)
[Use : $K_{b}$ for water $= 0.52 \, K \, kg \, mol^{-1}$,Boiling point of water $= 100^{\circ} C$]

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The molecular weight of benzoic acid in benzene as determined by the depression in freezing point method corresponds to:

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$A$ solute $A$ dimerizes in water. The boiling point of a $2 \, M$ solution of $A$ is $100.52^{\circ} C$. The percentage association of $A$ is ..... $\%$.(Round off to the Nearest integer)[Use : $K_{b}$ for water $= 0.52 \, K \, kg \, mol^{-1}$,Boiling point of water $= 100^{\circ} C$]

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The Van't Hoff factor $i$ for a $0.2 \ m$ aqueous solution of urea is

The molecular weight of benzoic acid in benzene as determined by the depression in freezing point method corresponds to:

During the determination of the molar mass of benzoic acid in benzene using the freezing point depression method,it corresponds to which of the following?

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$A$ solute $A$ dimerizes in water. The boiling point of a $2 \, M$ solution of $A$ is $100.52^{\circ} C$. The percentage association of $A$ is ..... $\%$.
(Round off to the Nearest integer)
[Use : $K_{b}$ for water $= 0.52 \, K \, kg \, mol^{-1}$,Boiling point of water $= 100^{\circ} C$]