Question

Pls answer my doubt I am sending an attachment.
Q. No 33.

Answer 1

The idea here is that since freezing-point depression is a colligative property, it will depend exclusively on how many particles of solute are present in solution, and not on the nature of those particles.

Tf = i . Kf . b

In your case, the solutions are equimolal, which means that they have equal molalities.

Since the solvent is the same in both cases, you can conclude that the difference between the freezing-point depression of the solution that contains potassium chloride,  KCl , and the solution that contains  X , will depend on the van't Hoff factors.

Since every mole of potassium chloride that dissolves in solution produces one mole of potassium cations and one mole of chloride anions, the van't Hoff factor for KCl is 2

This means that the van't Hoff factor of compound  X  will be

(1/4) x 2 = 0.5

So, when you dissolve one mole of compound  

X

in water, you get  

0.5

moles of particles of solute.

This should tell you that compound  X  does not dissociate in aqueous solution, nor does it hydrolyze, since these two processes would create more moles of solute particles per mole of solute dissolved.

If  50 %  of the molecules of  X  dimerize, you will end up with 3n/4 moles of solute finally

On the other hand, if  75 %  of the molecules of  X  were to form trimers, which are compound that contain three monomers, you'd have a final of n/2 moles of solute

Therefore, the answer is (3) Trimerise to extent of  75 %