Question

A sample of liquid $H_{2}O$ of mass 18.0 g injected into an evacuated 7.6 L flask maintained at ${27.0}^o$C .If vapour pressure of $H_{2}O$ at ${27}^0$C is 24.63 mm Hg. What weight percentage of the water will be vaporized when the system comes to equilibrium ? Assume water vapours behave as an ideal gas . The volume occupied by the liquid water is negligible compared to the volume of the container ;

• A. 1%
• B. 10%
• C. 18%
• D. 20%

Answer 1

The correct option is A 1%

According to ideal gas equation, $PV=nRT$

Vapour pressure $\left(P\right)=24.63mmHg=\frac{24.63}{760}atm$

Volume occupied by Vapour(V)=7.6 L ( as volume occupies by liquid water is negligible)$T={276}^{o}C=27+273=300K$

So, moles of vapor, $n=PV/RT$

$n=\frac{24.63\times 7.6}{760\times 0.0821\times 300}=0.01mol$

So, mass of vapour $H_{2}O=0.01\times 18=0.18g$

So, 0.18 g of liquid water is vapourised.

% of water that vapourised $=\frac{0.18}{18}\times 100=1$%

Hence, correct answer is (a)