Question

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 ?
(Given : Vapour pressure of $H_2O$ at $300\, K$ is $3170\, Pa ;\, R = 8.314 \,J \,K^{-1}\, mol^{-1}$)

• A. $5.56 \times 10^{-3}\, mol$
• B. $1.53 \times 10^{-2}\, mol$
• C. $4.46 \times 10^{-2}\, mol$
• D. $1.27 \times 10^{-3}\, mol$

Answer 1

Answer: (D)

$n = \frac{PV}{RT}$
$= 128 \times 10^{-5}$ moles
$= \frac{3170 \times 10^{-5} atm \times 1L}{0.0821L\, atm\, k^{-1}\, mol^{-1}\times 300K} \approx 1.27 \times 10^{-3} \,mol$