The freezing point of a 1.00 m aqueous solut | vedclass

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(D) The depression in freezing point is given by $\Delta T_f = T_f^{\circ} - T_f = 0 - (-1.91) = 1.91 \ K$. Using the formula $\Delta T_f = i 	imes K

Vedclass · Accepted Answer

$2.7$

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(D) The depression in freezing point is given by $\Delta T_f = T_f^{\circ} - T_f = 0 - (-1.91) = 1.91 \ K$. Using the formula $\Delta T_f = i 	imes K_f 	imes m$,where $i$ is the van't Hoff factor,$K_f = 1.86 \ K \ kg \ mol^{-1}$,and $m = 1.00 \ m$: $i = \frac{\Delta T_f}{K_f 	imes m} = \frac{1.91}{1.86 	imes 1.00} \approx 1.027$. For the dissociation of $HF$: $HF ightleftharpoons H^+ + F^-$. If $\alpha$ is the degree of dissociation,then $i = 1 + \alpha$. $1 + \alpha = 1.027 \implies \alpha = 0.027$. Therefore,the percentage dissociation is $0.027 	imes 100 = 2.7 \%$.

The freezing point of a $1.00 \ m$ aqueous solution of $HF$ is found to be $-1.91 \ ^oC$. The freezing point constant of water,$K_f$ is $1.86 \ K \ kg \ mol^{-1}$. The percentage dissociation of $HF$ at this concentration is ................ $\%$

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