First slide

Gaseous state

Question

The volume of 0.0168 mole of 0₂ obtained by thermal decomposition of KClO₃ and collected by displacement of water is 428 mL at a pressure of 754 mmHg at 25°C. The
pressure of water Vapour at 25°C is

Moderate

A

18 mm Hg
B

20 mm Hg
C

24 mm Hg
D

30 mm Hg

Solution

pV = nRT
$p_{atm} = \frac{n}{V} RT = \frac{0 .0168 \times 0 .0821 \times 298}{0 .428} = 0 .9603 atm$
$p_{atm} = 730 mmHg ∴ p_{total} = p_{atm} + p_{vapour} 754 = 730 + p_{vapour} V_{vapour} = 24 mmHg$

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