150 g of acetic acid was contaminated with 10.2 g ascorbic acid (C6H8O6) to lower down its freezing point by (x ×10-1)C o. The value of x is _________________.(Nearest integer) [Given Kf kg mol-1.,Molar mass of ascorbic acid = 176 g mol-1]

150 g CH3 COOH 10.2g ascorbic acid ⇒0.058 moles ∆Tf=x×10-1C o ∆Tf=kf·molality x×10-1C o=3.9×0.058150×1000 x×10-1C o=1.5C o x×10-1C o=15×10-1C o

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150 g of acetic acid was contaminated with 10.2 g ascorbic acid (C₆H₈O₆) to lower down its freezing point by (x $\times$ 10^-1) ${}^{o}C .$ The value of x is _________________.
(Nearest integer) [Given K_f kg mol^-1.,
Molar mass of ascorbic acid = 176 g mol^-1]

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answer is 15.

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Detailed Solution

$150 g C H_{3} C O O H 10.2 g a s c o r b i c a c i d \Rightarrow 0.058 m o l e s ∆ T_{f} = (x \times 10^{- 1}) {}^{o}C ∆ T_{f} = k_{f} \cdot m o l a l i t y (x \times 10^{- 1}) {}^{o}C = 3.9 \times \frac{0.058}{150} \times 1000 (x \times 10^{- 1}) {}^{o}C = 1.5 {}^{o}C (x \times 10^{- 1}) {}^{o}C = 15 \times 10^{- 1} {}^{o}C$