$1.00 \, g$ of a non-electrolyte solute dissolved in $50 \, g$ of benzene lowered the freezing point of benzene by $0.40 \, K$. The freezing point depression constant of benzene is $5.12 \, K \, kg \, mol^{-1}$. Find the molar mass of the solute.
(N/A) The formula for molar mass $(M_2)$ using freezing point depression is:
$M_2 = \frac{K_f \times w_2 \times 1000}{\Delta T_f \times w_1}$
Given:
$K_f = 5.12 \, K \, kg \, mol^{-1}$
$w_2 = 1.00 \, g$
$w_1 = 50 \, g$
$\Delta T_f = 0.40 \, K$
Substituting the values:
$M_2 = \frac{5.12 \, K \, kg \, mol^{-1} \times 1.00 \, g \times 1000 \, g \, kg^{-1}}{0.40 \, K \times 50 \, g} = 256 \, g \, mol^{-1}$
Thus,the molar mass of the solute is $256 \, g \, mol^{-1}$.
$M_2 = \frac{K_f \times w_2 \times 1000}{\Delta T_f \times w_1}$
Given:
$K_f = 5.12 \, K \, kg \, mol^{-1}$
$w_2 = 1.00 \, g$
$w_1 = 50 \, g$
$\Delta T_f = 0.40 \, K$
Substituting the values:
$M_2 = \frac{5.12 \, K \, kg \, mol^{-1} \times 1.00 \, g \times 1000 \, g \, kg^{-1}}{0.40 \, K \times 50 \, g} = 256 \, g \, mol^{-1}$
Thus,the molar mass of the solute is $256 \, g \, mol^{-1}$.
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