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

Benzene and toluene form ideal solution over the entire range of compo

1.	Benzene and toluene form ideal solution over the entire range of composition. The vapour pressures of pure benzeneand toluene at 300 K are 50.71 mm Hg and 32.06 mm Hg respectively. Calculate the mole fraction of benzene in the vapour phase if 80 g of benzene is mixed with 100 g of toluene.
Answer» SOLUTION :Molar mass of benzene `(C_(6)H_(6))="78 g mol"^(-1),"Molar mass of toluene "(C_(6)H_(5)CH_(3))="92 g mol"^(-1)` `therefore"No. of moles in 80 g of benzene "=("80 g")/("78 g mol"^(-1))="1.026 mole"` `"No. of moles in 100 g of toluene "=("100 g")/("92 g mol"^(-1))="1.087 mole"` `therefore"In the solution, mole fraction of benzene "=(1.026)/(1.-26+1.087)=(1.026)/(2.113)=0.486` mole fraction of toluene `=1-0.486=0.514` `p_("Benzene")^(@)="50.71mm, "p_("Toluene")^(@)="32.06 mm"` `"Applying RAOULT's law,"p_("Benzene")=x_("Benzene")xxp_("Benzene")^(@)=0.486xx"50.71 mm = 24.65 mm"` `""p_("Total")=x_("Toluene")=0.514xx"32.06 mm = 16.48 mm"` `therefore"Mole fraction of benzene in the vapour phase"=(p_("Benzene"))/(p_("Benzene")+p_("Toluene")) =(24.65)/(24.65+16.48) =(24.65)/(41.13) = 0.60`