$1.00 ext{ g}$ of a non-electrolyte solute (molar mass $250 ext{ g/mol}$) was dissolved in $51.2 ext{ g}

Question

$1.00 	ext{ g}$ of a non-electrolyte solute (molar mass $250 	ext{ g/mol}$) was dissolved in $51.2 	ext{ g}$ of benzene. If the freezing point depression constant, $K_f$ of benzene is $5.12 	ext{ K kg mol}^{-1}$, the freezing point of benzene will be lowered by:

Accepted Answer

Given:
Mass of solute ($w_2$) = $1.00 	ext{ g}$
Molar mass of solute ($M_2$) = $250 	ext{ g/mol}$
Mass of solvent (benzene, $w_1$) = $51.2 	ext{ g} = 0.0512 	ext{ kg}$
Freezing point depression constant ($K_f$) = $5.12 	ext{ K kg mol}^{-1}$

The depression in freezing point ($\Delta T_f$) is given by the formula:
$\Delta T_f = K_f 	imes m$
Where $m$ is the molality of the solution.
$m = \frac{w_2}{M_2 	imes w_1 	ext{ (in kg)}} = \frac{1.00}{250 	imes 0.0512} = \frac{1}{12.8} 	ext{ m}$

Now, substitute the values into the equation:
$\Delta T_f = 5.12 	imes \frac{1}{12.8} = 0.4 	ext{ K}$

Therefore, the freezing point of benzene will be lowered by $0.4 	ext{ K}$.

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