If $10^{-4} \mathrm{dm}^3$ of water is introduced into a $1.0 \mathrm{dm}^3$ flask to $300 \mathrm{~K}$, how…

If $10^{-4} \mathrm{dm}^3$ of water is introduced into a $1.0 \mathrm{dm}^3$ flask to $300 \mathrm{~K}$, how many moles of water are in the vapour phase when equilibrium is established? (Given : Vapour pressure of $\mathrm{H}_2 \mathrm{O}$ at $300 \mathrm{~K}$ is $3170 \mathrm{~Pa} ; \mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ )
  1. $5.56 \times 10^{-3} \mathrm{~mol}$
  2. $1.53 \times 10^{-2} \mathrm{~mol}$
  3. $4.46 \times 10^{-2} \mathrm{~mol}$
  4. $1.27 \times 10^{-3} \mathrm{~mol}$

Solution

$\mathrm{n}=\frac{\mathrm{PV}}{\mathrm{RT}}=$ $=128 \times 10^{-5} \mathrm{moles}$ $=\frac{3170 \times 10^{-5} \mathrm{~atm} \times 1 \mathrm{~L}}{0.0821 \mathrm{~L} \mathrm{~atm} \mathrm{\textrm {k } ^ { - 1 } \mathrm { mol } ^ { - 1 } \times 3 0 0 \mathrm { K }}} \approx 1.27 \times 10^{-3} \mathrm{~mol}$

Asked in: JEE Main 2010

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