Question

Two solutions of A and B are available. The first is known to contain 1 mole of A and 3 moles of B and its total vapour pressure is 1. Atm. The second is known to contain 2 moles of A and 2 moles of B; its vapour pressure is greater than 1 atm, but it is found that this total vapour pressure may be reduced to 1 atm by the addition of 6 moles of C. the vapour pressure of pure C is 0.80 atm. Assuming ideal solutions and that all these data refer to 25C, calculate the vapour pressureof pure A and of pure B.

Answer 1

solution $1^{st}$

$n_A=1mole;n_B=3mole$

$\therefore x_A=\frac{1}{1+3}=0.25$

& $x_B=\frac{3}{1+3}=0.75$

$\therefore$ Raolt's Law

$P=P_A^o{x}_A+P_B^o{x}_B$

$\Rightarrow 1=0.25P_A^o+0.75P_B^o....\left(1\right)$

solution $2^{nd}$;

$x_A=\frac{2}{2+2+6}=0.2;x_B=\frac{2}{2+2+6}=0.2;x_c=\frac{6}{2+2+6}=0.6$

$\therefore$ Raolt's Law

$1=0.20P_A^o+0.20P_B^o+0.60P_C^o.......\left(2\right)$

$\Rightarrow 1=0.20P_A^o+0.20P_B^o+0.60\times 0.8$

$\Rightarrow 0.52=0.20(P_A^o+P_B^o)$

$\Rightarrow P_A^o+P_B^o=\mathrm{2.6......}\left(3\right)$

From $\left(1\right)$ & $\left(3\right)$

we get $P_A^o=1.9atm$

& $P_B^o=0.7atm$