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(D) The depression in freezing point $(\Delta T_f)$ is given by the formula: $\Delta T_f = K_f \cdot m$,where $m$ is the molality of the solution.
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Vedclass · Accepted Answer

$M_2 = \frac{1000 \cdot K_f \cdot W_2}{\Delta T_f \cdot W_1}$

Vedclass · Answer

(D) The depression in freezing point $(\Delta T_f)$ is given by the formula: $\Delta T_f = K_f \cdot m$,where $m$ is the molality of the solution.
Molality $(m)$ is defined as the number of moles of solute $(n_2)$ per kilogram of solvent $(W_1 	ext{ in grams})$: $m = \frac{W_2 \cdot 1000}{M_2 \cdot W_1}$.
Substituting this into the depression in freezing point equation: $\Delta T_f = K_f \cdot \frac{W_2 \cdot 1000}{M_2 \cdot W_1}$.
Rearranging the formula to solve for the molar mass of the solute $(M_2)$: $M_2 = \frac{1000 \cdot K_f \cdot W_2}{\Delta T_f \cdot W_1}$.

What is the relation between the depression in freezing point and the molar mass of a non-volatile solute?

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