Question

When dissolved in 100 g of solvent whose molecular weight is 94.10 and whose freezing point is 45 ºC, 0.5550 g of solute of molecular weight 110.1 gave in freezing point of depression of 0.382 ºC. Again, the 0.4372 g of solute of unknown molecular weight has dissolved in 96.50 g of solvent, freezing point lowering was found to be 0.467 ºC . Find the molecular weight of the unknown state.

Answer 1

is the freezing point of depression.

k_f is the freezing point depression constant which depends on the solvent

m is the the molality of the solution.

Case 1)

  = 0.382^oC

For molality,

number of moles of solute,

mass of solvent,

Now solving the expression for k_f,

k_f= 7578.054 ^oC/g of solvent

Case (2)

= 0.467 ºC

k_f= 7578.054 ^oC/g of solvent

Substituting in the equation,

Molecular weight, M=73.52

Hope this is clear.

Have a nice day;)