Question

determine the molality of a solution having 0.2 mole fraction of I2 present in C6H6​

Answer 1

hey mate here ur answer :-

ANSWER

Let the solution be x molal.

Let the solution be x molal.Moles of benzene present in 1000g of benzene =

Let the solution be x molal.Moles of benzene present in 1000g of benzene = 78

Let the solution be x molal.Moles of benzene present in 1000g of benzene = 781000

Let the solution be x molal.Moles of benzene present in 1000g of benzene = 781000

Let the solution be x molal.Moles of benzene present in 1000g of benzene = 781000 =120.82.

Let the solution be x molal.Moles of benzene present in 1000g of benzene = 781000 =120.82.Mole fraction of solute =

Let the solution be x molal.Moles of benzene present in 1000g of benzene = 781000 =120.82.Mole fraction of solute = x+12.82

Let the solution be x molal.Moles of benzene present in 1000g of benzene = 781000 =120.82.Mole fraction of solute = x+12.82x=0.2=

Let the solution be x molal.Moles of benzene present in 1000g of benzene = 781000 =120.82.Mole fraction of solute = x+12.82x=0.2= x+12.82

Let the solution be x molal.Moles of benzene present in 1000g of benzene = 781000 =120.82.Mole fraction of solute = x+12.82x=0.2= x+12.82x

Let the solution be x molal.Moles of benzene present in 1000g of benzene = 781000 =120.82.Mole fraction of solute = x+12.82x=0.2= x+12.82x

Let the solution be x molal.Moles of benzene present in 1000g of benzene = 781000 =120.82.Mole fraction of solute = x+12.82x=0.2= x+12.82x .

Let the solution be x molal.Moles of benzene present in 1000g of benzene = 781000 =120.82.Mole fraction of solute = x+12.82x=0.2= x+12.82x .On solving for x we get x=3.2