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States of Matter
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07/09/2021 11:10 am
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The volume of 0.0168 mol of O2 obtained by decomposition of KClO3 and collected by displacement of water is 428 mL at a pressure of 754 mm Hg at 25 °C. The pressure of water vapour at 25°C is
(a) 18 mm Hg
(b) 20 mm Hg
(c) 22 mm Hg
(d) 24 mm Hg
1 Answer
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07/09/2021 11:16 am
Correct answer: (d) 24 mm Hg
Explanation:
Volume of 0.0168 moles at STP
= 0.0168 × 22400 = 376.3mL
\(\frac{P_1V_1}{T_1}\) = \(\frac{P_2V_2}{T_2}\)
or \(\frac{760 \times 376.3}{273}\) = \(\frac{P_2 \times 428}{298}\)
or P2 = 730 mm
Pressure of water = 754 - 730 = 24 mm Hg