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Abnormal Colligative Properties

0.971 mole of...

Question

0.971 mole of $N_{2} O_{4}$ and 0.0580 moles of $N O_{2}$ are in equilibrium in a flask of 0.750 litres at $27^{0} C$ . This flask is connected to an evacuated flask of 2.25 litre at $27^{0} C$ . A new equilibrium is established. Assuming the volume of connecting tube negligible, if the new molar concentrations at new equilibrium are $X$ M and $Y$ M.

The value of $10000 (X + Y)$ is :

3523

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5647

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6168

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2879

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Solution

The correct option is D 3523
$N_{2} O_{4} ⇌ 2 N O_{2}$

At equilibrium, $[N_{2} O_{4}] = \frac{0.971}{0.750} = 1.2946$
At equilibrium, $[N O_{2}] = \frac{0.0580}{0.750} = 0.0773$

$K = \frac{[N O_{2}]^{2}}{[N_{2} O_{4}]}$

$K = \frac{(0.0773)^{2}}{1.2946} = 4.619 \times 10^{- 3}$

The total volume becomes $V = 2.25 + 0.750 = 3$ after connecting to evacuated flask.

$[N_{2} O_{4}] = \frac{1.2946 \times 0.75}{3} - a = 0.32365 - a$

$[N O_{2}] = \frac{0.073 \times 0.750}{3} = 0.0193 + 2 a$

$K = \frac{(0.0193 + 2 a)^{2}}{0.32365 - a} = 4.619 \times 10^{- 3}$

$a = 0.0091$

$X = [N_{2} O_{4}] = 0.0193 + 2 (0.009) = 0.0375$

$Y = [N O_{2}] = 0.32365 - 0.0091 = 0.3175$

$10000 (X + Y) = 10000 (0.0375 + 0.3175) = 3523$

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