# Equimolal solutions of "KCl" and compound "X" in "H"_2"O" show depressions of freezing pts. in a 4:1 ratio . Assuming "KCl" to be ionized completely , the compound "X" must ?

## A) dissociate to extent of 50 % B) Hydrolyse to the extent of 80 % C) Trimerize to extent of 75 % D) Dimerize to extent of 50%

May 18, 2016

Here's my take on this.

#### Explanation:

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.

The equation that allows you to calculate freezing-point depression looks like this

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \Delta {T}_{f} = i \cdot {K}_{f} \cdot b \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$, where

$\Delta {T}_{f}$ - the freezing-point depression;
$i$ - the van't Hoff factor
${K}_{f}$ - the cryoscopic constant of the solvent;
$b$ - the molality of the solution.

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, $\text{KCl}$, and the solution that contains $\text{X}$, will depend on the van't Hoff factors.

You will have

$\Delta {T}_{\text{f KCl" = i_"KCl}} \cdot {K}_{f} \cdot b$

$\Delta {T}_{\text{f X" = i_"X}} \cdot {K}_{f} \cdot b$

Divide these two equations to get

$\left(\Delta {T}_{\text{f KCl")/(DeltaT_"f X") = (i_"KCl" * color(red)(cancel(color(black)(K_f * b))))/(i_"f X}} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{{K}_{f} \cdot b}}}\right)$

But since you know that the freezing-point depressions are in a $4 : 1$ ratio, you can say that

(DeltaT_"f KCl")/(DeltaT_"f X") = i_"f KCl"/i_"f X" = 4/1

This is equivalent to

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{i}_{\text{f X" = 1/4 xx i_"f KCl}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Now, the van't Hoff factors tells you the ratio that exists between the number of moles of solute dissolved in solution and the number of moles of particles of solute that are created in solution when the solute dissolves.

Potassium chloride dissociates to form potassium cations, ${\text{K}}^{+}$, and chloride anions, ${\text{Cl}}^{-}$

${\text{KCl"_ ((aq)) -> "K"_ ((aq))^(+) + "Cl}}_{\left(a q\right)}^{-}$

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 this compound will be

${i}_{\text{f KCl}} = 2$

This means that the van't Hoff factor of compound $\text{X}$ will be

${i}_{\text{f X}} = \frac{1}{4} \times 2 = 0.5$

So, when you dissolve one mole of compound $\text{X}$ in water, you get $0.5$ moles of particles of solute.

Right from the start, this should tell you that compound $\text{X}$ 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.

Now, let's assume that you dissolve $n$ moles of compound $\text{X}$ in solution. You need these moles of solute to become $\frac{n}{2}$ moles of particles.

A dimer is formed when two monomers, which in your case would be two molecules of $\text{X}$, are bonded together.

If 50% of the molecules of $\text{X}$ dimerize, you will end up with

$\frac{n}{2} \to$ remain as molecules of $\text{X}$

$\frac{1}{2} \cdot \frac{n}{2} = \frac{n}{4} \to$ exist as dimers

In this case, $n$ moles of $\text{X}$ would create

$\frac{n}{2} + \frac{n}{4} = \frac{3 n}{4} \to$ moles of particles of solute

Not what we're looking for. On the other hand, if 75% of the molecules of $\text{X}$ were to form trimers, which are compound that contain three monomers, you'd have

$\frac{n}{4} \to$ remain as molecules of $\text{X}$

$\frac{1}{3} \cdot \frac{3 n}{4} = \frac{n}{4} \to$ exist as trimers

In this case, $n$ moles of $\text{X}$ would create

$\frac{n}{4} + \frac{n}{4} = \frac{n}{2} \to$ moles of particles of solute

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