Home
Class 12
CHEMISTRY
The mole fraction of toluene in the vapo...

The mole fraction of toluene in the vapour phase which is in equilibrium with a solution of benzene `(P_(B)^(@)=120 "torr")` and toluene `(P_(T)^(@)=80 "torr")` having `2.0 mol` of each, is

A

0.5

B

0.25

C

0.6

D

0.4

Text Solution

Verified by Experts

The correct Answer is:
D
`chi'_(T)(vap) =(chi_(T)P_(T)^(@))/(chi_(T)P_(T)^(@) + chi_(B)P_(B)^(@))`
`=0.4`
Promotional Banner

Similar Questions

Explore conceptually related problems

Total Vapour pressure of mixture of 1mol A (p_(A)^(0)=150 "torr") and 2mol B (p_(B)^(0)=240 "torr") is 200 "torr" . In this case

Ethylene bromide C_(2)H_(4)Br_(2) , and 1,2 -dibromopropane, C_(3)H_(6)Br_(2) ,form a series of ideal solutions over the whole range of composition. At 85^(@)C , the vapour pressure of these two pure liquids are 173 and 127 torr, respectively. . If 10.0 g of ethylene bromide is dissolved in 80.0 g of 1,2-dibromopropane, calculate the partial pressure of each component and teh total pressure of the solution at 85^(@)C . b. Calculate the mole fraction of ethylene bromide in the vapour in equilibrium with the above solution. c. What would be the mole fraction of ethylene bromide in a solution at 85^(@)C equilibrated with a 50 : 50 mole mixture in the vapour?

What is the composition of the vapour which is in equilibrium at 30@C with a benzene-toluene solution with a mole fraction of benzene of (a) 0.400 and (b) 0.600 ? P_(b)@ = 119 torr , P_(t)@ = 37.0 torr

1 mol benzene (P^(@)_("benzene")=42 mm) and 2 mol toluence (P^(@)_("toluene")=36 mm) will have

(a) The vapour pressures of benzene and toluene at 293 Kare 75 mm Hg and 22 mm Hg respectively. 23.4 g of benzene and 64.4 g of toluene are mixed. If the two form an ideal solution, calculate the mole fraction of benzene in the vapour phase assuming that the vapour pressures are in equilibrium with the liquid mixture at this temperature. (b) What is meant by +ve and -ve deviations from Raoult's law and how is the sign of H solution related to +ve and -ve deviations from Raoult's law ?

Mole fraction of component A in vapour phase is chi_(1) and that of component A in liquid mixture is chi_2 , then ( p_(A)^@ )= vapour pressure of pure A, p_(B)^@ = vapour pressure of pure B), the total vapour pressure of liquid mixture is

Mole fraction of component A in vapour phase is chi_(1) and that of component A in liquid mixture is chi_2 , then ( p_(A)^@ )= vapour pressure of pure A, p_(B)^@ = vapour pressure of pure B), the total vapour pressure of liquid mixture is

The vapour pressure of pure components A and B are 200 torr and 100 torr respectively. Assuming a solution of these components obeys Raoult's law, the mole fraction of component. A in vapour phase in equilibrium with a solution containing equimoles of A and B is :

Two liquids A and B have vapour pressures in the ratio of p_(A)@,p_(B)@ = 1 : 2 at a certain temperature. Suppose we have an ideal solution of A and B in the mole fraction ratio A:B = 1:2 . What would be the mole fraction of A in the vapour in equilibrium with the solution at a given temperature? a. 0.25 , b. 0.2 , c. 0.5 d. 0.33

A certain ideal solution of two liquids A and B has mole fraction of 0.3 and 0.5 for the vapour phase and liquid phase, respectively. What would be the mole fraction of B in the vapour phase, when the mole fraction of A in the liquid is 0.25 ?