At 300K an ideal solution is formed by mixing 460g of toluene with 390g of benzene.If the vapour pressure of pure toluene and benzene at 300k are 32 and 40mm respectively the mole fraction of toluene in vapour phase is

Benzene and toluene form ideal solution over the entire range of composition. The vapour pressure of pure benzene and naphthalene at 300K are 50.71 mm Hg and 32.06mm Hg , respectively. Calculate the mole fraction of benzene in vapour phase if 80g of benzene is mixed with 100g of naphthalene.

Equal moles of benzene and toluene are mixed. The vapour pressure of benzene and toluene in pure state are 700 and 600 mm Hg respectively. The mole fraction of benzene in vapour state is :-

Benzene and naphthalene form ideal solution over the entire range of composition. The vapour pressure of pure benzene and naphthalene at 300 K are 50.71 mm Hg and 32.06 mm Hg respectively. Calculate the mole fraction of benzene in vapour phase if 80 g of benzene is mixed with 100 g of naphthalene.

A solution has 1:4 mole ratio of pentane to hexane . The vapour pressure of pure hydrocarbons at 20^@C are 440 mm Hg for pentane and 120 mm Hg for hexane .The mole fraction of pentane in the vapour phase is

The vapour pressure of ethanol and methanol are 44.0 mm and 88.0 mm Hg , respectively. An ideal solution is formed at the same temperature by mixing 60 g of ethanol with 40g of methanol. Calculate the total vapour pressure of the solution and the mole fraction of methanol in the vapour.

The vapour pressure of mixture of toluene and xylene at 90^(@) C is 0.5 atm. If at this temperature 400 mm and 150 mm respectively then what will be the mole fraction of toluene in mixture ?

The vapour pressures of ethanol and methanol are 44.5 and 88.7 mm Hg , respectively. An ideal solution is formed at the same temperature by mixing 60 g of ethanol with 40 g of methanol . Calculate the total vapour pressure of the solution and mole fraction of methanol in the vapour.

The vapour pressures of ethanol and methanol are 44.5 and 88.7 mm Hg , respectively. An ideal solution is formed at the same temperature by mixing 60 g of ethanol with 40 g of methanol . Calculate the total vapour pressure of the solution and mole fraction of methanol in the vapour.

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

Benzene (C_(6)H_(6), 78 g//mol) and toluene (C_(7)H_(8),92g//mol) form an ideal solution. At 60^(@)C the vapour pressure of pure benzene and pure toluene are 0.507 atm and 0.184 atm, respectively. The mole fraction of benezen in a solution of these two chemicals that has a vapour preesure of 0.350 atm at 60^(@)C , will be :