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

Step-by-step solution

Given, (w_a=460 ~g ) (toluene) (w_B=390 ~g (benzene) ) Vapour pressure of pure toluene ( (p_A )=32 ~mm ) Vapour pressure of pure benzene ( (p_B )=40 ~mm ) Moles of toluene, ( (n_A )= w_A M_A = 460 92 ) Moles of benzene, (n_B= w_B M_B = 390 78 ) Hence, mole fraction of toluene ( _A ). ( _A= 460 92 460 92 + 390 78 ) Mole fraction of benzene ( ( _B ) ) ( gathered _B= 390 78 460 92 + 390 78 p_ total =p^ _A _A+p^ _B _B p_ total =32 ( 460 92 460 92 + 390 78 )+40 ( 390 78 390 78 + 460 92 ) p_ total =36 ~mm gathered ) Mole