You visited us 4 times! Enjoying our articles? Unlock Full Access!

Raoult's Law

The vapour pr...

Question

The vapour pressure of pure liquids 1 and 2 are $550 m m H g$ and $800 m m H g$ respectively at $350 K$ .
What will be the mole fraction of component 2 in liquid phase when total vapour pressure of the solution is $600 m m H g$ ?

0.1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.2

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

0.9

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.8

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $0.2$
By Raoult's Law :
$P_{T} = p_{1}^{\circ} χ_{1} + p_{2}^{\circ} χ_{2}$
where
$P_{T}$ is the total pressure of the solution.
$p_{1}^{\circ}$ is the vapour pressure of pure component 1
$p_{2}^{\circ}$ is the vapour pressure of pure component 2
$χ_{1}$ is the mole fraction of component 1 in liquid phase.
$χ_{2}$ is the mole fraction of component 2 in liquid phase.

Putting the values:
$600 = 550 χ_{1} + 800 χ_{2} 600 = 550 χ_{1} + 800 (1 - χ_{1}) 600 = 800 - 250 χ_{1} χ_{1} = 0.8$
$χ_{2} = 1 - 0.8 = 0.2$

Suggest Corrections

Similar questions

Q. The vapour pressure of pure liquids 1 and 2 are

450 m m H g

and

700 m m H g

respectively at

350 K

.
When total vapour pressureof the solution is

600 m m H g

, the mole fraction of component 1 in liquid phase will be:

Q. The vapour pressure of pure liquids 1 and 2 are

450 m m H g

and

700 m m H g

respectively at

350 K

. At a point when total vapour pressure is

600 m m H g

, choose the correct option(s) :

Q. Consider a binary mixture of two ideal volatile liquids A and B having vapour pressure of

300 m m H g

and

600 m m H g

in their pure state.
What will be the mole fraction of component A in vapour phase

(y_{A})

if the total pressure

(P_{T})

of the solution is

400 m m H g

Q. In a binary mixture of two ideal volatile liquids A and B, calculate the mole fraction of component A in vapour phase if:
Vapour pressure of A in pure state is

500 m m H g

Vapour pressure of B in pure state is

700 m m H g

Total pressure

(P_{T})

of the solution is

600 m m H g

The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K. Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase.

Join BYJU'S Learning Program

Explore more

Raoult's Law

Standard XII Chemistry

Join BYJU'S Learning Program

The vapour pressure of pure liquids 1 and 2 are 550 mm Hg and 800 mm Hg respectively at 350 K. What will be the mole fraction of component 2 in liquid phase when total vapour pressure of the solution is 600 mm Hg?

The vapour pressure of pure liquids 1 and 2 are $550 m m H g$ and $800 m m H g$ respectively at $350 K$ .
What will be the mole fraction of component 2 in liquid phase when total vapour pressure of the solution is $600 m m H g$ ?