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Relative Lowering of Vapour Pressure

X moles of ...

Question

X moles of $P C l_{5}$ must be added to one-litre vessel at $250$ K to obtain a concentration of 0.1 moles of $C l_{2}$ .

$K_{c}$ for, $P C l_{5} ⇌ P C l_{3} + C l_{2}$ , is $0.0414$ mol / litre.

The value of 1000X is ____.

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Solution

$P C l_{5} \to P C l_{3} + C l_{2}$
At $t = 0$ $x$ $0$ $0$
At equilibrium $x - y$ $y$ $y$

But $y = 0.1$ (Given)

$K_{c} = \frac{0.1 \times 0.1}{x - 0.1}$

$= 0.0414$

$x - 0.1 = \frac{1}{4.14}$

$x = 0.1 + \frac{1}{4.14}$

$= 0.3415$ moles

Here we get $X = 0.3415 m o l e s$

So $1000 X = 341 m o l e s$

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P C l_{5}

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