JEE Main 2022 · 28 Jul · Shift 1 · Chemistry Q84

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ChemistryInteger+4 / -1Question 84 of 90

150 g of acetic acid was contaminated with 10.2 g ascorbic acid $(\mathrm{C_6H_8O_6})$ to lower down its freezing point by $(x\times10^{-1})$ °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 $^{-1}$ ]

Your answer

Answer and worked solution

Official answer: 15

Molality $\frac{10.2/176}{0.150}=0.386$ , so $\Delta T_f=3.9\times0.386=1.51$ °C $\approx15\times10^{-1}$ .

Answer: 15

Topics:Colligative properties

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JEE Main 2022 · 28 Jul · Shift 1, Chemistry question 84