Benzene and toluene form ideal solution over the entire range of composition. The vapour pressure of pure benzene and toluene 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 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 pressure of pure benzene and toluene 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 toluene.
Answer» Solution :MOLAR mass of `(C_(6)H_(5))` `= 6xx12+6xx1` `= 78 g mol^(-1)` Molar mass of toluene `(C_(6)H_(5)CH_(3))` `= 7xx12+8xx1` `= 92 g mol^(-1)` Now, no. of moles present in 80 g of benzene `= (80)/(78)=1.026` mol And, no. of moles present in 100 g of toluene `= (100)/(92)=0.087` mol Therefore, Mole FRACTION of benzene, `x_(b)=(1.026)/(1.026+1.087)` = 0.486 And, mole fraction of toluene, `x_(t)=1-0.486` = 0.514 It is given that vapour presure of PURE benzene, `p_(b)^(0)=50.71` mm Hg Therefore, partial vapour presure of benzene, `p_(b)=x_(b)xx p_(b)` = 24.645 mm Hg And, partial vapour pressure of toluene, `p_(1)=x_(t)xx p_(t)` `= 0.514xx32.06` = 16.479 mm Hg Hence, mole fraction of benzene in vapour PHASE is given by : `=(p_(b))/(p_(b)+p_(t))` `= (24.645)/(24.645+16.479)` `= (24.645)/(41.124)` = 0.599 =0.6