Question

For a solution if $p_A^0=600mmHg$, $p_B^0=840mmHg$ under atmospheric conditions and vapour pressure of solution is 1 atm then find
i) Composition of solution
ii) Composition of vapour in equilibrium with solution.

Answer 1

(i) According to Raoult's law

$P_A\alpha x_A$

also $x_A+x_B=1$

$\Rightarrow P_A^0{x}_A+P_B^0{x}_B=P$

from question : $P=1$ atm

$P_A^0=600mm$ $Hg$

$P_B^0=840mm$ $Hg$

$760mm$ $Hg=1$ atm

$\Rightarrow 600mmHg=\frac{1}{760}\times 600atm=\frac{60}{76}$ atm

$\Rightarrow 840mmHg=\frac{84}{76}$ atm

So, $1=\frac{60}{76}(1-x_B)+\frac{84}{76}\left(x_B\right)$

$\Rightarrow x_B=\frac{16}{24}$

$x_B=2/3;x_A=1/3$

composition of vapour phase in equilibrium with solution is-

$y_A=\frac{P_A}{P_{total}}=\frac{P_A^0-x_A}{1}=\frac{P_A^0{x}_A}{1}=\frac{20}{76}=0.26$

$y_B=\frac{P_B}{P_{total}}=\frac{P_B^0{x}_B}{1}=\frac{28}{38}=0.74$

(ii) Relative lowerins of vapour pressure $=\frac{\Delta P}{P^0}=\frac{P_A^0-P_A}{P_A^0}=x_B=\frac{W_B{M}_A}{M_B{W}_A}$

vapour pressure of water $=360mm$ $Hg$

Volume of water $=200$ ml

mass $=200$g

$\frac{\Delta P}{P^0}=0.5=\frac{W_B\times 18}{200\times 60}$

$W_B=\frac{200\times 60\times 0.5}{18}=333.33$ grams.