Question

Two miscible liquids A and B having vapour pressure in pure state $P_A^o$ and $P_B^o$ are mixed in mole fraction ${\chi }_A$ and ${\chi }_B$ to get a mixture having total vapour pressure of mixture $P_M$.Which of the following reactions are correct?

• A. ${\chi }_A=\frac{P_M-P_B^o}{P_A^o-P_B^o}$
• B. $\frac{{\chi }_{A\left(l\right)}}{{\chi }_{A\left(V\right)}^{{}^{\prime }}}=\frac{P_M}{P_A^o}$
• C. $\frac{{\chi }_{A\left(l\right)}}{{\chi }_{A\left(V\right)}^{{}^{\prime }}}=\frac{P_M}{P_B^o}$
• D. All of these

Answer 1

Answer: (A), (B)

${\chi }_A=\frac{P_M-P_B^o}{P_A^o-P_B^o}$
$\frac{{\chi }_{A\left(l\right)}}{{\chi }_{A\left(V\right)}^{{}^{\prime }}}=\frac{P_M}{P_A^o}$

Using Roult's law for binary mixtures,

$P_M=P_A^o\cdot {\chi }_A+P_B^o\cdot {\chi }_B$;

Also,$P^{{}^{\prime }}=P_M\cdot {\chi }_{A\left(V\right)}=P_A^o\cdot {\chi }_{A\left(l\right)}$

So, $\frac{P_M}{P_A^o}=\frac{{\chi }_{A\left(l\right)}}{{\chi }_{A\left(v\right)}}$

$=P_A^o\cdot {\chi }_A+P_B^o\cdot (1-{\chi }_A)$

$\therefore {\chi }_A=\frac{P_M-P_B^o}{P_A^o-P_B^o}$

Hence, options A and B are correct.