Calculate the mass of ascorbic acid (Vitamin $C$,$C_{6}H_{8}O_{6}$) to be dissolved in $75 \ g$ of acetic acid to lower its melting point by $1.5 \ ^{\circ}C$. $K_{f} = 3.9 \ K \ kg \ mol^{-1}$.

(N/A) Mass of acetic acid,$w_{1} = 75 \ g$.
Molar mass of ascorbic acid $(C_{6}H_{8}O_{6})$,$M_{2} = (6 \times 12) + (8 \times 1) + (6 \times 16) = 176 \ g \ mol^{-1}$.
Lowering of melting point,$\Delta T_{f} = 1.5 \ K$.
Using the formula for depression in freezing point:
$\Delta T_{f} = \frac{K_{f} \times w_{2} \times 1000}{M_{2} \times w_{1}}$.
Rearranging for the mass of solute $(w_{2})$:
$w_{2} = \frac{\Delta T_{f} \times M_{2} \times w_{1}}{K_{f} \times 1000}$.
Substituting the values:
$w_{2} = \frac{1.5 \times 176 \times 75}{3.9 \times 1000} = \frac{19800}{3900} \approx 5.077 \ g$.
Hence,approximately $5.08 \ g$ of ascorbic acid is required.