Question

Assume that 0.10 moles of a certain substance is dissolved in 250. g of water. The freezing point of the resultant solution is -2.2 °C. I: Which of the following choices is the dissolved substance? Kf = 1.86°C/m for water.

a: C6H12O6

b: CO(NH2)2

c: NaCl

d: CaCl2

II: Calculate the % dissociation of the substance

I know the answer for (I) is D, and the answer for (II) is 98% but I am not sure how it is achieved.

Answer 1

Ans Part I . d) CaCl2

Delta Tf = i. kf . m

where delta Tf is the depression in freezing point

i is the vant hoff's factor

and m is the molality

So putting all the values we get :

2.2 = i . 1.86 x 0.10/ 0.250

i = 2.96

The vant hoff factor of 2.96 corresponds to the CaCl2 , hence d is the answer for part I

Part II :

CaCl2 = Ca²⁺ + 2Cl^-

So 1 mole of CaCl2 gives 3 moles of ions ,

since here the vant hoff factor calculated is 2.96 ,

So the % dissociation of the substance = ( 2.96 / 3) x 100

= 98%