Question

For a solution if p°a =600 mm hg, p°b =840 mm hg under atmospheric condition and vapour pressure of solution is 1 atmosphere then find •question 1 composition of solution •question 2 composition of vapour in equlibrium with solution​

Answer 1

Answer : The composition of solution $X_a$ and $X_b$ are, 0.34 and 0.66 respectively.

The composition of vapor in equilibrium with solution $Y_a$ and $Y_b$ are, 0.27 and 0.73 respectively.

Explanation :

(a) First we have to calculate the composition of solution.

According to the Rouault's law,

$p_s=p_a+p_b$

$p_s=p^o_aX_a+p^o_bX_b$     ............(1)

where,

$p_s$ = vapor pressure of solution = 1 atm = 760 mmHg

$p^o_a$ = vapor pressure of pure 'a'= 600 mmHg

$p^o_b$ = vapor pressure of pure 'b'= 840 mmHg

$X_a$ and $X_b$ = mole of fraction of 'a' and 'b'

Now put all the given values in equation 1, we get :

$760=600X_a+840X_b$      .............(2)

And as we know that,

$X_a+X_b=1$         ...............(3)

Now equating equation 2 and 3, we get

$X_a=0.34$

$X_b=0.66$

Therefore, the composition of solution $X_a$ and $X_b$ are, 0.34 and 0.66 respectively.

(b) Now we have to calculate the vapor pressure of 'a' and 'b'.

$p_a=p^o_a\times X_a=600\times 0.34=204mmHg$

$p_b=p^o_b\times X_b=840\times 0.66=554.4mmHg$

Now we have to calculate the composition of vapor in equilibrium with solution​.

$Y_a=\frac{p_a}{p_a+p_b}=\frac{204}{204+554.4}=0.27$

$Y_b=\frac{p_b}{p_a+p_b}=\frac{554.4}{204+554.4}=0.73$

Therefore, the composition of vapor in equilibrium with solution $Y_a$ and $Y_b$ are, 0.27 and 0.73 respectively.

Answer 2

Explanation:

detailed solution of the question in the attached photo.