Question

25 grams of a compound with a formula weight of 110g/mol was dissolved in 500mL of water, and the freezing point of the aqueous solution decreased to -1.69°C. What does the formula of the compound look like? The freezing point depression constant of water is 1.86°C/m, and you may assume that the density of water is 1.00g/mL.

a. The compound is molecular

b. Ionic compound with formula AX

c. Ionic compound with formula AX2 or A2X (could be either)

d. Ionic compound with formula AX3, A2X2, or A3X (could be either)

e. None of these is correct

Answer 1

Ans b) Ionic compound with formula AX

Freezing point depression = i. kf . m

where i is the vant hoff's factor

kf is the freezing point constant = 1.86 ^oC/m

m is the molality

m = (25 / 110 ) / 0.5

m = 0.454 m

Now putting all the values in the formula , we get :

1.69 = i x 1.86 x 0.454

i = 2

Now since the vant hoff's factor of the compound comes out to be 2 , that means the compound molecule has completely dissociated into 2 ions . Hence it is an ionic compound with the formula AX.