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150 \mathrm{~g} of acetic acid was contaminated with 10.2 \mathrm{~g} ascorbic acid \left(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{6}\right) to lower down its freezing point by \left(x \times 10^{-1}\right)^{\circ} \mathrm{C}. The value of \mathrm{x} is___________.(Nearest integer)
[Given \mathrm{K}_{f}=3.9 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1};
molar mass of ascorbic acid =176 \mathrm{~g} \mathrm{~mol}^{-1}]

Option: 1

15


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

Given , 150 g of \mathrm{CH_{3}COOH}
10.2 g of ascorbic acod having 0.058 moles.

\mathrm{K_{f}= 3.9\, K\, kg\, mol^{-1}}.
M. mass of ascorbic acid = 176 g/mol.
 \mathrm{\text{no.of moles of ascorbic acid}= \frac{10.2}{176}= 0.058\, moles.}

\mathrm{\Delta T_{f}= K_{f}.m\; \; \; m= molality}
           \mathrm{= 3.9\times \frac{0.058}{150}\times 1000}
           \mathrm{= 1.5^{\circ}C}
           \mathrm{= \left ( 15\times 10^{-1} \right )^{\circ}C}

Hence answer is 15
   

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