Question

# A solution of 12.00 grams of an unknown nonelectrolyte compound is dissolved in 200.0 grams of...

A solution of 12.00 grams of an unknown nonelectrolyte compound is dissolved in 200.0 grams of benzene. The resultant solution freezes at 3.45oC. What is the molar mass of the unknown compound? [The freezing point of pure benzene is 5.45oC ; the Kf for benzene is 5.07oC 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

Given data:

Wt. of unknown non-electrolyte w= 12.00g

Wt. of solvent Benzene W = 200g

Freezing point of benzene = 5.45oC

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

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

Cryscopic constant, Kf = 5.07oC/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, Kf, Delta Tf as,

m = (1000 x Kf x w) / (Delta Tf= 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)

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