JEE MainChemistrySolutions
125 ~g of a solute X (molar mass = 100 ~g ~mol ⁻¹ ) is dissolved in 400 ~g of a solvent Y (molar mass = 80 ~g ~mol ⁻¹ ). The solute undergoes partial dimerization in the solvent. The observed depression in freezing point of the solution is 10 ~K . If the relative lowering of vapour pressure of the solution is 1 x , the value of x is _ _ _ _ . [Given: Freezing point depression constant of the solvent K_f = 4.0 ~K ~kg
Correct answer
6
Step-by-step solution
Moles of solute X initially taken: n_X = 125 100 = 1.25 ~mol Moles of solvent Y : n_Y = 400 80 = 5.0 ~mol Molality of the solution ( m ) based on initial moles: m = 1.25 0.400 = 3.125 ~mol ~kg ⁻¹ The theoretical depression in freezing point is: T_ f( theoretical ) = K_f m = 4.0 3.125 = 12.5 ~K The van 't Hoff factor ( i ) is the ratio of observed to theoretical depression: i = T_ f( observed ) T_ f( theoretical ) = 10 12.5 = 0.8 Effective moles of solute particles in the solution: n_ effective = i n_X = 0.8 1.25 =