Of the following four aqueous solutions, the total number of those solutions whose freezing point is lower than that of 0.10 M C$$_2$$H$$_5$$OH is _________ (Integer answer)
(i) 0.10 M Ba$$_3$$(PO$$_4$$)$$_2$$
(ii) 0.10 M Na$$_2$$SO$$_4$$
(iii) 0.10 M KCl
(iv) 0.10 M Li$$_3$$PO$$_4$$

We recall the colligative-property formula for depression in freezing point:

$$\Delta T_f = i\,K_f\,m$$

Here $$\Delta T_f$$ is the lowering of the freezing point, $$i$$ is the van’t Hoff factor (total number of particles formed in solution per formula unit), $$K_f$$ is the cryoscopic constant of the solvent (water in this case), and $$m$$ is the molality of the solution. For all the solutions given, the concentration is stated as 0.10 M; at such low concentration the numerical value of molarity and molality are practically the same, so we can safely put $$m = 0.10\ \text{mol kg}^{-1}$$ for every solution.

First, let us evaluate the reference solution, 0.10 M ethanol, C$$_2$$H$$_5$$OH. Ethanol is a nonelectrolyte; it does not dissociate into ions. Therefore

$$i_{\text{EtOH}} = 1$$

Substituting in the formula gives

$$\Delta T_{f,\text{EtOH}} = 1 \times K_f \times 0.10 = 0.10\,K_f$$

Any solution that produces a greater $$\Delta T_f$$ (that is, any solution for which $$i > 1$$ under the same molality) will have its freezing point lower than that of the ethanol solution. Now we examine each salt one by one, writing its ionic dissociation equation, counting the ions, finding $$i$$, and then comparing its $$\Delta T_f$$ with $$0.10\,K_f$$.

(i) For barium phosphate, Ba$$_3$$(PO$$_4$$)$$_2$$:

$$\text{Ba}_3(\text{PO}_4)_2 \rightarrow 3\,\text{Ba}^{2+} + 2\,\text{PO}_4^{3-}$$

Number of ions = $$3 + 2 = 5$$, hence

$$i_{\,\text{Ba}_3(\text{PO}_4)_2} = 5$$

Therefore

$$\Delta T_{f,\,\text{Ba}_3(\text{PO}_4)_2} = 5 \times K_f \times 0.10 = 0.50\,K_f$$

Clearly $$0.50\,K_f > 0.10\,K_f$$, so this solution freezes lower than the ethanol solution.

(ii) For sodium sulfate, Na$$_2$$SO$$_4$$:

$$\text{Na}_2\text{SO}_4 \rightarrow 2\,\text{Na}^{+} + \text{SO}_4^{2-}$$

Number of ions = $$2 + 1 = 3$$, thus

$$i_{\text{Na}_2\text{SO}_4} = 3$$

Hence

$$\Delta T_{f,\,\text{Na}_2\text{SO}_4} = 3 \times K_f \times 0.10 = 0.30\,K_f$$

Again $$0.30\,K_f > 0.10\,K_f$$, so this solution also freezes lower.

(iii) For potassium chloride, KCl:

$$\text{KCl} \rightarrow \text{K}^{+} + \text{Cl}^{-}$$

Number of ions = $$1 + 1 = 2$$, giving

$$i_{\text{KCl}} = 2$$

Thus

$$\Delta T_{f,\,\text{KCl}} = 2 \times K_f \times 0.10 = 0.20\,K_f$$

Since $$0.20\,K_f > 0.10\,K_f$$, this solution also has a lower freezing point than ethanol.

(iv) For lithium phosphate, Li$$_3$$PO$$_4$$:

$$\text{Li}_3\text{PO}_4 \rightarrow 3\,\text{Li}^{+} + \text{PO}_4^{3-}$$

Number of ions = $$3 + 1 = 4$$, so

$$i_{\text{Li}_3\text{PO}_4} = 4$$

Hence

$$\Delta T_{f,\,\text{Li}_3\text{PO}_4} = 4 \times K_f \times 0.10 = 0.40\,K_f$$

Again $$0.40\,K_f > 0.10\,K_f$$, so this one too freezes lower.

We observe that for every one of the four salts the van’t Hoff factor is greater than 1, which makes their calculated $$\Delta T_f$$ larger than $$0.10\,K_f$$, the value for the 0.10 M ethanol reference. Consequently, the freezing points of all four salt solutions are lower than that of the 0.10 M ethanol solution.

So, the total number of such solutions is $$4$$.

Hence, the correct answer is Option 4.