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09/09/2021 7:27 pm
Topic starter
If 10–4 dm3 of water is introduced into a 1.0 dm3 flask at 300 K, how many moles of water are in the vapour phase when equilibrium is established? (Given: Vapour pressure of H2O at 300 K is 3170 Pa; R = 8.314 J K–1 mol–1)
(a) 5.56 × 10–3 mol
(b) 1.53 × 10–2 mol
(c) 4.46 × 10–2 mol
(d) 1.27 × 10–3 mol
1 Answer
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09/09/2021 7:31 pm
Correct answer: (d) 1.27 × 10–3 mol
Explanation:
From the ideal gas equation :
PV = nRT
or n = \(\frac{PV}{RT}\)
= \(\frac{3170 \times 10^{-3}}{8.314 \times 300}\)
= 1.27 x 10–3 mol