If 10^-4 dm^3 of water is introduced into a 1.0 dm^3 flask to 300 K , how many moles of water are in the vapour phase when equilibrium is established? - Chemistry MCQ AIEEE 2026 | ExamTest.live

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If 1 0^{- 4} dm^{3} of water is introduced into a 1.0 dm^{3} flask to 300 K, how many moles of water are in the vapour phase when equilibrium is established?(Given: Vapour pressure of H_{2} O at 300 K is 3170 Pa; R = 8.314 J K^{- 1} mol^{- 1})

Solution & Step-by-step Explanation

Using the ideal gas equation P V = n RT : P = 3170 Pa V = 1.0 dm^{3} = 1 0^{- 3} m^{3} T = 300 K R = 8.314 J K^{- 1} mol^{- 1} $ n = \frac{P V}{RT} $ n = \frac{3170 \times 1 0 ^{- 3}}{8.314 \times 300} n = \frac{3.170}{2494.2} \approx 1.27 \times 1 0^{- 3} mol Check if liquid water remains: 1 0^{- 4} dm^{3} = 0.1 mL \approx 0.1 g .Moles of water introduced = \frac{0.1}{18} \approx 5.5 \times 1 0^{- 3} mol .Since 1.27 \times 1 0^{- 3} < 5.5 \times 1 0^{- 3}, liquid water is present, and equilibrium vapour pressure is reached.

Try it yourself before checking the explanation above.

If 1 0^{- 4} dm^{3} of water is introduced into a 1.0 dm^{3} flask to 300 K, how many moles of water are in the vapour phase when equilibrium is established?(Given: Vapour pressure of H_{2} O at 300 K is 3170 Pa; R = 8.314 J K^{- 1} mol^{- 1})

5.56 \times 1 0^{- 3} mol

1.53 \times 1 0^{- 2} mol

4.46 \times 1 0^{- 2} mol

1.27 \times 1 0^{- 3} mol

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