One molal solution of a carboxylic acid in be | vedclass

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(D) The formula for elevation of boiling point is $\Delta T_{b} = i 	imes K_{b} 	imes m$. Given $\Delta T_{b} = 1.518 \ K$,$K_{b} = 2.53 \ K \ kg \

Vedclass · Accepted Answer

$80$

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(D) The formula for elevation of boiling point is $\Delta T_{b} = i 	imes K_{b} 	imes m$. Given $\Delta T_{b} = 1.518 \ K$,$K_{b} = 2.53 \ K \ kg \ mol^{-1}$,and $m = 1 \ mol \ kg^{-1}$. For dimerization,the van't Hoff factor is $i = 1 - \alpha + \frac{\alpha}{2} = 1 - \frac{\alpha}{2}$. Substituting the values: $1.518 = (1 - \frac{\alpha}{2}) 	imes 2.53 	imes 1$. $\frac{1.518}{2.53} = 1 - \frac{\alpha}{2}$. $0.6 = 1 - \frac{\alpha}{2}$. $\frac{\alpha}{2} = 1 - 0.6 = 0.4$. $\alpha = 0.8$,which is $80 \%$.

One molal solution of a carboxylic acid in benzene shows the elevation of boiling point of $1.518 \ K$. The degree of association for dimerization of the acid in benzene is $........ \%$. ($K_b$ for benzene $= 2.53 \ K \ kg \ mol^{-1}$)

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