The freezing point of an equimolal aqueous so | vedclass

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(D) The depression in freezing point is given by the formula $\Delta T_f = i 	imes K_f 	imes m$. For equimolal solutions,$\Delta T_f$ is directly pr

Vedclass · Accepted Answer

$C_6H_{12}O_6$

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(D) The depression in freezing point is given by the formula $\Delta T_f = i 	imes K_f 	imes m$. For equimolal solutions,$\Delta T_f$ is directly proportional to the van't Hoff factor $(i)$. $C_6H_{12}O_6$ (glucose) is a non-electrolyte,so its $i = 1$. $C_6H_5NH_3^+Cl^-$ dissociates into $2$ ions $(i = 2)$. $Ca(NO_3)_2$ dissociates into $3$ ions $(i = 3)$. $La(NO_3)_3$ dissociates into $4$ ions $(i = 4)$. Since glucose has the lowest $i$ value,it will have the minimum depression in freezing point $(\Delta T_f)$. Therefore,the freezing point $(T_f = T_f^0 - \Delta T_f)$ will be the highest for $C_6H_{12}O_6$.

The freezing point of an equimolal aqueous solution will be highest for:

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