An aqueous solution of a weak monobasic acid containing $0.1 ext{ g}$ in $21.7 ext{ g}$ of water freezes a

Question

An aqueous solution of a weak monobasic acid containing $0.1 	ext{ g}$ in $21.7 	ext{ g}$ of water freezes at $272.817 	ext{ K}$. If the value of $K_f$ for water is $1.86 	ext{ K kg mol}^{-1}$, the molecular mass of the acid is:

Accepted Answer

The molecular mass of the solute ($M_2$) can be calculated using the formula for depression of freezing point:
$$M_2 = \frac{1000 	imes w_2 	imes K_f}{\Delta T_f 	imes w_1}$$

Where:
*   $w_2$ = mass of solute = $0.1 	ext{ g}$
*   $w_1$ = mass of solvent = $21.7 	ext{ g}$
*   $K_f$ = molal freezing point depression constant = $1.86 	ext{ K kg mol}^{-1}$
*   $\Delta T_f$ = depression in freezing point = $T_f^0 - T_f$
    *   To match the options, we assume the freezing point of pure water ($T_f^0$) is taken as $273 	ext{ K}$ (an approximation often used in older problems).
    *   $\Delta T_f = 273 	ext{ K} - 272.817 	ext{ K} = 0.183 	ext{ K}$

Substituting the values:
$$M_2 = \frac{1000 	imes 0.1 	imes 1.86}{0.183 	imes 21.7}$$
$$M_2 = \frac{186}{3.9711}$$
$$M_2 \approx 46.83 	ext{ g mol}^{-1}$$

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