In steam distillation of toluene, the pressure of toluene in vapour is:

equal pressure of barometer

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

less than pressure of barometer

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

equal to vapour pressure to toluene in simple distillation

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

more than vapour pressure of toluene in simple distillation

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

Open in App

Solution

The correct option is A less than pressure of barometer

In steam distillation of toluene, the pressure of toluene in vapour is less than the pressure of barometer, because it is carried out when a solid or liquid is insoluble in water and is volatile with steam but the impurities are non-volatile.

Option B is correct.

Suggest Corrections

Similar questions

The vapour pressure of pure benzene and toluene are 160 and 60 torr respectively. The molefraction of toluene in vapour phase in contact with equimolar solution of benzene and toluene is

The vapour pressure of a benzene – toluene binary solution at certain temperature is given by $P = 450 - 100 X_{T}$ .
Where $X_{T}$ is the mole fraction of toluene. Find the vapour pressures of Benzene and Toluene in pure state.

Q. At

88^{0} C

, benzene has a vapour pressure of 900 torr and toluene has a vapour pressure of 360 torr. What is the mole fraction of benzene in the mixture with toluene that will boil at

88^{0} C

at 1 atm pressure, benzene -toluene from an ideal solution?

Q. The vapour pressure of pure benzene and toluene are 160 and 60 torr respectively. The mole fraction of toluene in the vapour phase in contact with an equimolar solution of benzene and toluene is :

Q. At

88^{0} C

benzene has a vapour pressure of 900 torr and toluene has a vapour pressure of 360 torr. What is the mole fraction of benzene in the mixture with toluene that will boil at

88^{0} C

at 1 atm pressure?

Join BYJU'S Learning Program