A solute A dimerizes in water. The boiling po | vedclass

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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

Vedclass · Accepted Answer

$100$

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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 $\Delta T_{b} = i 	imes K_{b} 	imes m$,where $m = 2 \, M$:$0.52 = (1 - \frac{\alpha}{2}) 	imes 0.52 	imes 2$.Dividing both sides by $0.52$,we get $1 = (1 - \frac{\alpha}{2}) 	imes 2$.$1 = 2 - \alpha$.$\alpha = 1$.Thus,the percentage association is $1 	imes 100 = 100 \, \%$.

$A$ 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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$A$ 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 $= 100^{\circ} C$]