Question

Two liquids a and b are mixed in 1:4 mole ratio to form an ideal solution if pure liquid a an b exert a vapour pressure of 75 mmhg and 22 mmhg, the vapour pressure of liquid a in vapour phase would be

Answer 1

the vapour pressure of a in vapour phase is praportional to its mole fraction
= 75(1/5)
= 15mmHg

and net vapour pressure of liquid is
= 15 + 22(4/5)
= 15 + 17.6
= 32.6 mmHg

Answer 2

Answer: The vapor pressure of liquid a in vapor phase would be 34.5 mm Hg

Explanation:  

According to Raoult's law, the vapor pressure of a component at a given temperature is equal to the mole fraction of that component multiplied by the vapor pressure of that component in the pure state.

$p_a=x_ap_a^0$ and $p_b=x_bP_b^0$

where, x = mole fraction

$p^0$ = pressure in the pure state

According to Dalton's law, the total pressure is the sum of individual pressures.

$p_{total}=p_a+p_b$

$p_{total}=x_ap_a^0+x_bP_b^0$

given $x_{a}=\frac{n_a}{n_a+n_b}=\frac{1}{1+4}=0.2$

$x_b=\frac{n_b}{n_a+n_b}=\frac{4}{1+4}=0.8$

$p_a^0=75mmHg$

$p_b^0=22mmHg$

$p_{total}=0.2\times 75+0.8\times 22=32.6 mmHg$

The mole fraction of liquid a  in vapor phase is given by:

$y_a=\frac{p_a}{P_{total}}$  

$p_a=x_ap_a^0=0.2\times 75mmHg=15mmHg$

$y_a=\frac{15mmHg}{32.6mmHg}=0.46$

Thus the vapor pressure of liquid a in vapor phase would be:

$p_a=y_a\times p_a^0=0.46\times 75mmHg=34.5mmHg$