Question

If 10^–4 dm³ of water is introduced into a 1.0 dm³ flask to 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)

(1) 5.56 x 10^–3 mol 

(2) 1.53 x 10^–2 mol 

(3) 4.46 x 10^–2 mol 

(4) 1.27 x 10^–3 mol

Answer 1

(4)

n = PV/RT = 

= 128 x 10^-5 moles

= (3770 x 10^-5atm x 1L/0.0821 L atm k^-1mol^-1 x 300K) = 1.27 x 10^-3 mol