Question

If ${10}^{-4}d{m}^3$ of water is introduced into a $1d{m}^3$ flask at 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/mol)

• A. $5.56\times {10}^{-6}mol$
• B. $1.53\times {10}^{-2}mol$
• C. $4.46\times {10}^{-2}mol$
• D. $1.27\times {10}^{-3}mol$

Answer 1

The correct option is D

$1.27\times {10}^{-3}mol$

$n=\frac{PV}{RT}=\frac{3170\times {10}^{-3}}{8.314\times 300}=1.27\times {10}^{-3}mol$