2.0 g of a non-electrolyte dissolved in 100 | vedclass

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(B) The formula for freezing point depression is $\Delta T_f = K_f 	imes m$, where $m$ is the molality. Molality $m = \frac{W_2 	imes 1000}{M_2 	im

Vedclass · Accepted Answer

$85 \ g \ mol^{-1}$

Vedclass · Answer

(B) The formula for freezing point depression is $\Delta T_f = K_f 	imes m$, where $m$ is the molality. Molality $m = \frac{W_2 	imes 1000}{M_2 	imes W_1}$, where $W_2$ is the mass of solute, $M_2$ is the molar mass of solute, and $W_1$ is the mass of solvent. Substituting the given values: $1.2 = 5.12 	imes \frac{2.0 	imes 1000}{M_2 	imes 100}$. $1.2 = \frac{5.12 	imes 20}{M_2}$. $M_2 = \frac{102.4}{1.2} \approx 85.33 \ g \ mol^{-1}$. Rounding to the nearest integer, the molar mass is $85 \ g \ mol^{-1}$.

$2.0 \ g$ of a non-electrolyte dissolved in $100 \ g$ of benzene lowers the freezing point of benzene by $1.2 \ K$. The freezing point depression constant of benzene is $5.12 \ K \ kg \ mol^{-1}$. The molar mass of the solute is:

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