Question

Four equimolal solutions of urea, NaCl, Ca(NO$_{3}$)$_{2}$ and Na$_{3}$PO$_{4}$ are prepared by dissolving them in water. Select the correct statement(s) assuming 100% ionisation of strong electrolyte from the following.

• A. Urea solution has lowest boiling point and highest freezing point
• B. Elevation in boiling point for Ca(NO$_{3}$)$_{2}$ solution is three times that of urea solution
• C. Depression in freezing point for NaCl solution is half to that for Na$_{3}$PO$_{4}$ solution
• D. Na$_{3}$PO$_{4}$ solution has highest freezing point and lowest boiling point

Answer 1

Answer: (A), (B), (C)

A, B, C

Answer 2

To determine the correct statements, we need to analyze the colligative properties of the given equimolal solutions. Colligative properties depend on the number of solute particles in the solution, which for electrolytes is represented by the van't Hoff factor (i).

The formulas for elevation in boiling point ($\Delta T_b$) and depression in freezing point ($\Delta T_f$) are:

$$\Delta T_b = i \cdot K_b \cdot m$$ $$\Delta T_f = i \cdot K_f \cdot m$$

Where $K_b$ is the molal elevation constant, $K_f$ is the molal depression constant, and $m$ is the molality of the solution. Since all solutions are equimolal, $m$ is the same for all. $K_b$ and $K_f$ are constants for the solvent (water).

The boiling point of the solution ($T_b$) is given by $T_b = T_b^0 + \Delta T_b$, where $T_b^0$ is the boiling point of the pure solvent. The freezing point of the solution ($T_f$) is given by $T_f = T_f^0 - \Delta T_f$, where $T_f^0$ is the freezing point of the pure solvent.

First, let's determine the van't Hoff factor (i) for each solute, assuming 100% ionization for strong electrolytes:

1. Urea (CO(NH$_{2}$)$_{2}$): Urea is a non-electrolyte and does not dissociate. $i = 1$

2. NaCl: NaCl is a strong electrolyte and dissociates into two ions (Na$^{+}$ and Cl$^{-}$). NaCl(aq) $\rightarrow$ Na$^{+}$(aq) + Cl$^{-}$(aq) $i = 2$

3. Ca(NO$_{3}$)$_{2}$: Ca(NO$_{3}$)$_{2}$ is a strong electrolyte and dissociates into three ions (Ca$^{2+}$ and 2NO$_{3}^{-}$). Ca(NO$_{3}$)$_{2}$(aq) $\rightarrow$ Ca$^{2+}$(aq) + 2NO$_{3}^{-}$(aq) $i = 3$

4. Na$_{3}$PO$_{4}$: Na$_{3}$PO$_{4}$ is a strong electrolyte and dissociates into four ions (3Na$^{+}$ and PO$_{4}^{3-}$). Na$_{3}$PO$_{4}$(aq) $\rightarrow$ 3Na$^{+}$(aq) + PO$_{4}^{3-}$(aq) $i = 4$

Now, let's compare the colligative properties based on the 'i' values: Since $\Delta T_b \propto i$ and $\Delta T_f \propto i$:

• A higher 'i' value leads to a greater elevation in boiling point ($\Delta T_b$) and thus a higher boiling point ($T_b$).
• A higher 'i' value leads to a greater depression in freezing point ($\Delta T_f$) and thus a lower freezing point ($T_f$).

The order of 'i' values is: Urea (1) < NaCl (2) < Ca(NO$_{3}$)$_{2}$ (3) < Na$_{3}$PO$_{4}$ (4)

Therefore:

• The order of $\Delta T_b$ and $T_b$ (from lowest to highest) is: Urea < NaCl < Ca(NO$_{3}$)$_{2}$ < Na$_{3}$PO$_{4}$ (Urea solution has the lowest boiling point; Na$_{3}$PO$_{4}$ solution has the highest boiling point).

• The order of $\Delta T_f$ (from lowest to highest) is: Urea < NaCl < Ca(NO$_{3}$)$_{2}$ < Na$_{3}$PO$_{4}$

• The order of $T_f$ (from highest to lowest) is: Urea > NaCl > Ca(NO$_{3}$)$_{2}$ > Na$_{3}$PO$_{4}$ (Urea solution has the highest freezing point; Na$_{3}$PO$_{4}$ solution has the lowest freezing point).

Now, let's evaluate each statement:

(A) Urea solution has lowest boiling point and highest freezing point

• Urea has the lowest 'i' value (i=1). This means it has the lowest $\Delta T_b$ and thus the lowest boiling point.
• It also has the lowest $\Delta T_f$ and thus the highest freezing point.
• This statement is Correct.

(B) Elevation in boiling point for Ca(NO$_{3}$)$_{2}$ solution is three times that of urea solution

• $\Delta T_b$ (Ca(NO$_{3}$)$_{2}$) = $i_{Ca(NO_{3})_{2}} \cdot K_b \cdot m = 3 \cdot K_b \cdot m$
• $\Delta T_b$ (Urea) = $i_{Urea} \cdot K_b \cdot m = 1 \cdot K_b \cdot m$
• Therefore, $\Delta T_b$ (Ca(NO$_{3}$)$_{2}$) = 3 $\times$ $\Delta T_b$ (Urea).
• This statement is Correct.

(C) Depression in freezing point for NaCl solution is half to that for Na$_{3}$PO$_{4}$ solution

• $\Delta T_f$ (NaCl) = $i_{NaCl} \cdot K_f \cdot m = 2 \cdot K_f \cdot m$
• $\Delta T_f$ (Na$_{3}$PO$_{4}$) = $i_{Na_{3}PO_{4}} \cdot K_f \cdot m = 4 \cdot K_f \cdot m$
• Therefore, $\Delta T_f$ (NaCl) = $(2/4) \cdot \Delta T_f$ (Na$_{3}$PO$_{4}$) = $0.5 \cdot \Delta T_f$ (Na$_{3}$PO$_{4}$).
• This statement is Correct.

(D) Na$_{3}$PO$_{4}$ solution has highest freezing point and lowest boiling point

• Na$_{3}$PO$_{4}$ has the highest 'i' value (i=4). This means it has the highest $\Delta T_b$ and thus the highest boiling point.
• It also has the highest $\Delta T_f$ and thus the lowest freezing point.
• The statement claims highest freezing point and lowest boiling point, which contradicts our findings.
• This statement is Incorrect.

Based on the analysis, statements (A), (B), and (C) are correct.