Question

two liquid having vapour pressure p1 and p2 in pure state in the ratio of 2:1 are mixed in the molar ratio of 1:2.the ratio of their moles in the vapor state would be

Answer 1

Vapour pressure of liquid 1, P1 = 2units
Vapour pressure of liquid 2, P2 = 1unit
Moles of liquid 1 = 1
Moles of liquid 2 = 2
Mole fraction of liquid 1 = 1/3
Mole fraction of liquid 2 = 2/3
Using Raoult's law:
Partial pressure of liquid 1 = Vapour pressure of liquid 1 mole freaction of liquid 1
Partial pressure of liquid 1 = 2 (1/3) = 2/3

Partial pressure of liquid 2 = ​Vapour pressure of liquid 2 mole freaction of liquid 2
Partial pressure of liquid 2 = 1​(2/3) = 2/3

Total pressure p = partial pressure of 1 + partial pressure of 2 = 2/3+2/3=4/3

Mole fraction of liquid 1 in vapour phase = partial pressure of liquid1/total vapour pressure =( 2/3)/(4/3)=1/2

Mole fraction of liquid 2 in vapour phase = partial pressure of liquid 2/ total Vapour pressure =(2/3)/(4/3)=1/2

Ratio of moles = 1:1

Answer 2

Answer:The ratio of their moles in the vapor state would be 1:1.

Explanation:

$p^0_1:p^o_2=2:1$

$\chi_1:\chi_2=1:2$

From above we can write as $n_1:n_2=2:1$

Mole fraction of two liquid in vapor phase.

$y_1=\frac{p_1}{p_1+p_2}$

$y_2=\frac{p_2}{p_1+p_2}$

$\frac{y_1}{y_2}=\frac{p_1}{p_2}$

According Dalton's law of partial pressure:

$p_1=\chi_1\times p^o_{1}$

$p_2=\chi_2\times p^o_{2}$

$\frac{y_1}{y_2}=\frac{\chi_1\times p^o_{1}}{\chi_2\times p^o_{2}}$

$\frac{y_1}{y_2}=\frac{n_1\times 2p^o_{2}}{2n_1\times 1p^o_{2}}$

$\frac{y_1}{y_2}=\frac{1}{1}$

The ratio of their moles in the vapor state would be 1:1.