Question

A solution of 12.00 grams of an unknown nonelectrolyte compound is dissolved in 200.0 grams of benzene. The resultant solution freezes at 3.45^oC. What is the molar mass of the unknown compound? [The freezing point of pure benzene is 5.45^oC ; the K_f for benzene is 5.07^oC m^-1]. (Hint: this is a two-step challenge, involving first finding the number of moles of solute in the solution from the freezing point data and then its molar mass in units of grams/mole)

A. 125 grams per mole

B. 152 grams per mole

C. 88.2 grams per mole

D. 55.8 grams per mole

E. 193 grams per mole

Answer 1

Given data:

Wt. of unknown non-electrolyte w= 12.00g

Wt. of solvent Benzene W = 200g

Freezing point of benzene = 5.45^oC

Freezing point of Benzene+ non-electrolyte solution= 3.45^oC

So, Depression in Freezing point DeltaT_f = 5.45 – 3.45 = 2.00^oC

Cryscopic constant, K_f = 5.07^oC/m

Molar mass of solute m = ? ( Or first moles (n) and then Molar mass m)

We know the relation between molality of solute (m) and w, W, K_f, Delta T_f as,

m = (1000 x K_f x w) / (Delta T_f= x W)

= (1000 x 5.07 x 12) / (2.00 x 200)

=152.1

= 152 (approx.)

Hence molar mass of solute is 152 grams per mole.

(Note: Direct formula used instead two steps that first to find moles and then molar mass)