Question

Of the following four aqueous solutions, total number of those solutions whose freezing point is lower than that of 0.10 M \(C_2H_5OH\) is _________. (Integer answer)
(i) 0.10 M \(Ba_3(PO_4)_2\)
(ii) 0.10 M \(Na_2SO_4\)
(iii) 0.10 M KCl
(iv) 0.10 M \(Li_3PO_4\)

Hint:

For equal molar concentrations of electrolytes and non-electrolytes, the freezing point is always lowest for the electrolyte with the highest number of ions (\(i\)).

Answer 1

Correct Answer: 4

Step 1: Understanding the Concept:
Freezing point depression (\(\Delta T_f\)) is a colligative property, meaning it depends on the number of particles in the solution. The freezing point of a solution is lower than the pure solvent. A higher concentration of particles leads to a lower freezing point.
Step 2: Key Formula or Approach:
\[ \Delta T_f = i \times K_f \times m \]
Since concentrations are given in Molarity (\(M\)) and are equal (\(0.10 \text{ M}\)), we can compare the values of the van't Hoff factor (\(i\)). Freezing point \(\propto - (i \times M)\).
Step 3: Detailed Explanation:
Reference solution: 0.10 M \(C_2H_5OH\) (ethanol). Ethanol is a non-electrolyte, so \(i = 1\).
Effective concentration of particles = \(1 \times 0.10 = 0.10 \text{ M}\).

Now consider the given solutions (assuming 100% dissociation):
(i) \(Ba_3(PO_4)_2 \rightarrow 3Ba^{2+} + 2PO_4^{3-}\). \(i = 5\). Effective conc. = \(0.50 \text{ M}\).
(ii) \(Na_2SO_4 \rightarrow 2Na^+ + SO_4^{2-}\). \(i = 3\). Effective conc. = \(0.30 \text{ M}\).
(iii) \(KCl \rightarrow K^+ + Cl^-\). \(i = 2\). Effective conc. = \(0.20 \text{ M}\).
(iv) \(Li_3PO_4 \rightarrow 3Li^+ + PO_4^{3-}\). \(i = 4\). Effective conc. = \(0.40 \text{ M}\).

All four solutions have a higher effective particle concentration (\(> 0.10 \text{ M}\)) than the reference ethanol solution. Therefore, all four will exhibit a greater freezing point depression, resulting in lower freezing points.
Step 4: Final Answer:
The total number of solutions is 4.