Question

At 80°vapourphase of benzene and toulene is 900and360 respectively find mole fraction of benzene on mixing toulene boil at 1atm

Answer 1

HEYA GM ❤

★ P°BENZENE = 900

P°BENZENE = 900 ★ P°TOLUENE = 360

P°BENZENE = 900 ★ P°TOLUENE = 360 ★ Mole fraction= ?

P°BENZENE = 900 ★ P°TOLUENE = 360 ★ Mole fraction= ? P(total) = [ P1° - P2° ] X¹ - P2°

P°BENZENE = 900 ★ P°TOLUENE = 360 ★ Mole fraction= ? P(total) = [ P1° - P2° ] X¹ - P2°760 = (900 - 360 ) X¹ - 360

P°BENZENE = 900 ★ P°TOLUENE = 360 ★ Mole fraction= ? P(total) = [ P1° - P2° ] X¹ - P2°760 = (900 - 360 ) X¹ - 360 760 - 360 = (540) X¹

P°BENZENE = 900 ★ P°TOLUENE = 360 ★ Mole fraction= ? P(total) = [ P1° - P2° ] X¹ - P2°760 = (900 - 360 ) X¹ - 360 760 - 360 = (540) X¹ X¹ = 400 / 540

★ X¹ ( Mole fraction ) = 0.74

X¹ ( Mole fraction ) = 0.74 TQ ❤ ❤