Solution of 0.78gmBenzene and 0.92gm of toluenev. p. of pure benzene =

1.	Solution of 0.78gmBenzene and 0.92gm of toluenev. p. of pure benzene = 400 mmHg.V.p. of pure toluene = 200 mm Hgmolar mass of Benzene = 78gmMolaremass of toluene= 92gmfind total v. p. of solution?
Answer» Answer: Apply Raoult law Explanation: TOTAL v.p = x1P1 + x2P2 where x1 is MOLE fraction of toluene P1 is v.p. of toluene x2 is mole fraction of BENZENE P2 is v.p. of toluene mole fraction = number of MOLES / total number of moles Moles = weight of substance / molar mass

Solution of 0.78gmBenzene and 0.92gm of toluenev. p. of pure benzene = 400 mmHg.V.p. of pure toluene = 200 mm Hgmolar mass of Benzene = 78gmMolaremass of toluene= 92gmfind total v. p. of solution?

Answer»

Answer:

Apply Raoult law

Explanation:

TOTAL v.p = x1P1 + x2P2

where x1 is MOLE fraction of toluene P1 is v.p. of toluene

x2 is mole fraction of BENZENE P2 is v.p. of toluene

mole fraction = number of MOLES / total number of moles

Moles = weight of substance / molar mass