150 g of acetic acid was contaminated with 10.2 g ascorbic acid $$(C_6H_8O_6)$$ to lower down its freezing point by $$x \times 10^{-1}$$ °C. The value of $$x$$ is _____ (Nearest integer). [Given $$K_f = 3.9$$ K kg mol$$^{-1}$$; Molar mass of ascorbic acid = 176 g mol$$^{-1}$$]

The mass of acetic acid (solvent) is 150 g = 0.150 kg, the mass of ascorbic acid ($$C_6H_8O_6$$, solute) is 10.2 g, the molar mass of ascorbic acid is 176 g/mol, and the cryoscopic constant $$K_f$$ of acetic acid is 3.9 K kg mol$$^{-1}$$.

Since the moles of ascorbic acid can be calculated by $$n = \frac{10.2}{176} = 0.05795 \text{ mol}$$, this represents the number of moles of solute present.

Substituting into the definition of molality gives $$m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.05795}{0.150} = 0.3864 \text{ mol/kg}$$.

Next, the depression in freezing point is given by $$\Delta T_f = K_f \times m$$. Substituting the values yields $$= 3.9 \times 0.3864$$ and thus $$= 1.507 \text{ °C}$$.

Expressing this in the required form gives $$\Delta T_f = 1.507 \text{ °C} = 15.07 \times 10^{-1} \text{ °C}$$. Therefore, the value of $$x$$ is 15 (nearest integer).