Solute A associates in water. When 0.7 g of | vedclass

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(D) The formula for freezing point depression is $\Delta T_f = i \cdot K_f \cdot m$.First,calculate the molality $(m)$: $m = \frac{0.7 \ g}{93 \ g \ m

Vedclass · Accepted Answer

$80$

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(D) The formula for freezing point depression is $\Delta T_f = i \cdot K_f \cdot m$.First,calculate the molality $(m)$: $m = \frac{0.7 \ g}{93 \ g \ mol^{-1}} 	imes \frac{1000 \ g \ kg^{-1}}{42.0 \ g} = 0.1792 \ mol \ kg^{-1}$.Using $\Delta T_f = 0.2 \ K$ and $K_f = 1.86 \ K \ kg \ mol^{-1}$,we find the van't Hoff factor $(i)$:$0.2 = i \cdot 1.86 \cdot 0.1792 \implies i = \frac{0.2}{1.86 \cdot 0.1792} \approx 0.60$.For the association reaction $2A ightleftharpoons A_2$,the van't Hoff factor is $i = 1 - \alpha + \frac{\alpha}{2} = 1 - \frac{\alpha}{2}$.Equating $1 - \frac{\alpha}{2} = 0.60$,we get $\frac{\alpha}{2} = 0.40$,so $\alpha = 0.80$.The percentage association is $80 \%$.

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