Question

mole fraction of I2 in benzene is 0.3 the molality of I2 in benzene is ( I = 127 )​

Answer 1

Answer:

5.5

Explanation:

We know that

Molality = $number of moles of solute * 1000/ weight of the solvent$(in grams)

Here iodine is solute and benzene is solvent

Given

Mole fraction of I₂ = 0.3

⇒ n( I₂ )/ [ n(I₂) + n(benzene)] = 0.3

Let n( I₂ ) = x and n( benzene) = y  [for our convinience]

⇒ x / (x+y) = 0.3

⇒ x = 0.3x + 0.3y

⇒ 0.7x = 0.3y

⇒ y = 7/3*x ________(1)

no. of moles of benzene = 7/3* no.of moles of iodine

We know that

no. of moles =  weight / Gram molecular weight

⇒no. of moles of benzene = y

⇒ weight of benzene / Gram molecular weight = y

We know that the molecular weight of benzene = 78

⇒ weight of benzene( solvent ) = 78*y

∴ Molality m = $number of moles of solute * 1000/ weight of the solvent$(in grams)

                    = x * 1000/weight of benzene

                    = x*1000 / 78*y

                    = 1000x / 78*(7/3*x)

                    = 1000x / 26*7*x

                    = 1000/ 182

                    = 5.4945 = 5.5(approx)