Question

At a particular temperature, the vapour pressures of two liquids A and B are respectively 120 and 180 mm of mercury. If 2 moles of A and 3 moles of B are mixed to form an ideal solution, the vapour pressure of the solution at the same temperature will be: (in mm of mercury)

• A. 156
• B. 145
• C. 150
• D. 108
• E. 48

Answer 1

Answer: (A)

$ P_{A}^{o}=120\,mm,P_{B}^{o}=180\,mm $
Number of moles of A and B = 2 moles and 3 moles respectively. Total pressure of solution
$ ={{P}_{A}}+{{P}_{B}} $
$ P=P_{A}^{o}{{x}_{A}}+P_{B}^{o}{{x}_{B}} $ ...(i)
where, $ x{}^\circ $ mole fraction, $ \left( \frac{n}{n+N} \right) $
$ x_{A}^{o}=\frac{2}{2+3}=\frac{2}{5}=0.4 $
$ x_{B}^{o}=\frac{2}{2+3}=\frac{3}{5}=0.6 $
On putting the values of $ x_{A}^{o},x_{B}^{o},p_{A}^{o} $
and $ p_{B}^{o} $ in Eq. (i)
$ =120\times 0.4+180\times 0.6 $
$ =48.0+108.0 $
$ =156.0\text{ }mm $