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)

$P_A^o=120mm,P_B^o=180mm$
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)
$=120\times 0.4+180\times 0.6$
$=48.0+108.0$
$=156.0mm$