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

JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Solutions

150 g of acetic acid was contaminated with 10.2 g ascorbic acid

(C_{6} H_{8} O_{6})

to lower down its freezing point by

(x \times 1 0^{- 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}

]

Answer: 15

Solution

Moles of ascorbic acid =

\frac{10.2}{176} = 0.058

mol. Molality =

\frac{0.058}{0.150} = 0.387

mol/kg. Depression in freezing point:

Δ T_{f} = K_{f} \times m = 3.9 \times 0.387 \approx 1. 5^{\circ} C = 15 \times 1 0^{- 1}

°C. Therefore, x = 15.

JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Integer answer

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