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Standard XII

Chemistry

Relative Lowering of Vapour Pressure

400 mL of amm...

Question

$400 m L$ of ammonia gas at a pressure of $300 m m$ and $200 m L$ of hydrogen chloride gas at a pressure of $600 m m$ are mixed in one litre evacuated flask. The pressure at normal temperature is :

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Solution

400 mL of ammonia gas at a pressure of 300 mm corresponds to

$n = \frac{P V}{R T} = \frac{300 \times 400}{R T} = \frac{120000}{R T}$
200 mL of hydrogen chloride gas at a pressure of 600 mm corresponds to

$n = \frac{P V}{R T} = \frac{200 \times 600}{R T} = \frac{120000}{R T}$
Thus an equal number of moles of ammonia and $H C l$ are present. They will react to form an equal number of moles of ammonium chloride.

Number of moles of ammonium chloride formed $= \frac{120000}{R T}$

The pressure at normal temperature is :

$P = \frac{n R T}{V}$

$P = \frac{\frac{120000}{R T} R T}{1000}$

$P = 120$ mm Hg.

Here volume 1 L is converted into 1000 mL.

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400 mL of ammonia gas at a pressure of 300 mm and 200 mL of hydrogen chloride gas at a pressure of 600 mm are mixed in one litre evacuated flask. The pressure at normal temperature is :

$400 m L$ of ammonia gas at a pressure of $300 m m$ and $200 m L$ of hydrogen chloride gas at a pressure of $600 m m$ are mixed in one litre evacuated flask. The pressure at normal temperature is :