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(C) Step $1$: Calculate the moles of ascorbic acid $(C_{6}H_{8}O_{6})$.$n = \frac{	ext{mass}}{	ext{molar mass}} = \frac{10.2 \ g}{176 \ g \ mol^{-1}

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$15$

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(C) Step $1$: Calculate the moles of ascorbic acid $(C_{6}H_{8}O_{6})$.$n = \frac{	ext{mass}}{	ext{molar mass}} = \frac{10.2 \ g}{176 \ g \ mol^{-1}} \approx 0.05795 \ mol$.Step $2$: Calculate the molality $(m)$ of the solution.$m = \frac{	ext{moles of solute}}{	ext{mass of solvent in kg}} = \frac{0.05795 \ mol}{0.150 \ kg} \approx 0.3863 \ mol \ kg^{-1}$.Step $3$: Calculate the depression in freezing point $(\Delta T_{f})$.$\Delta T_{f} = K_{f} 	imes m = 3.9 \ K \ kg \ mol^{-1} 	imes 0.3863 \ mol \ kg^{-1} \approx 1.5066^{\circ} C$.Step $4$: Express $\Delta T_{f}$ in the form $(x 	imes 10^{-1})^{\circ} C$.$1.5066 \approx 15.066 	imes 10^{-1}$.Rounding to the nearest integer,$x = 15$.

$150 \ g$ of acetic acid was contaminated with $10.2 \ g$ ascorbic acid $(C_{6}H_{8}O_{6})$ to lower its freezing point by $(x \times 10^{-1})^{\circ} C$. The value of $x$ is (Nearest integer). [Given $K_{f} = 3.9 \ K \ kg \ mol^{-1}$; Molar mass of ascorbic acid $= 176 \ g \ mol^{-1}$]

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