\(0.1 M\, \ NaCl\) solution has lower freezing point than \(0.1 M \, \ MgCl_2\) solution in water because:

van't Hoff factor for \(MgCl_2\) is more than \(NaCl\)

The correct answer is option 3. van't Hoff factor for \(MgCl_2\) is more than \(NaCl\).

Let us explore why a \(0.1 \text{ M} \, \text{NaCl}\) solution has a higher freezing point than a \(0.1 \text{ M} \, \text{MgCl}_2\) solution in water, focusing on the van’t Hoff factor and its effect on freezing point depression.

Freezing point depression (\( \Delta T_f \)) occurs when a solute is added to a solvent, causing the freezing point of the solution to decrease. The extent of this depression depends on the number of solute particles in the solution.

The formula for freezing point depression is:

\(\Delta T_f = i \cdot K_f \cdot m \)

where:

\( \Delta T_f \) is the decrease in freezing point

\( i \) is the van’t Hoff factor (number of particles the solute dissociates into)

\( K_f \) is the cryoscopic constant of the solvent (which is constant for a given solvent)

\( m \) is the molality of the solution (in this case, molarity can be used as an approximation because the solutions are dilute)

The van’t Hoff factor (\( i \)) represents the number of particles into which a solute dissociates in solution.

Sodium Chloride (NaCl)

NaCl dissociates into two ions in water: \( \text{Na}^+ \) and \( \text{Cl}^- \).

van’t Hoff Factor: For NaCl, \( i = 2 \).

When NaCl is dissolved, it produces 2 particles per formula unit of NaCl.

Magnesium Chloride (MgCl\(_2\))

MgCl\(_2\) dissociates into three ions in water: \( \text{Mg}^{2+} \) and 2 \( \text{Cl}^- \).

van’t Hoff Factor: For MgCl\(_2\), \( i = 3 \).

When MgCl\(_2\) is dissolved, it produces 3 particles per formula unit of MgCl\(_2\).

Comparing the Solutions

Molarity: Both solutions have the same molarity (0.1 M), meaning they have the same concentration of solute in terms of moles per liter.

van’t Hoff Factor: MgCl\(_2\) has a higher van’t Hoff factor (3) compared to NaCl (2). This means that for the same molarity, MgCl\(_2\) produces more particles in solution than NaCl.

Freezing Point Depression Calculation

For 0.1 M NaCl:

\( i = 2 \)

\( \Delta T_f = 2 \cdot K_f \cdot 0.1 \)

For 0.1 M MgCl\(_2\):

\( i = 3 \)

\( \Delta T_f = 3 \cdot K_f \cdot 0.1 \)

Since the van’t Hoff factor is greater for MgCl\(_2\), the freezing point depression (\( \Delta T_f \)) will be greater for the MgCl\(_2\) solution compared to the NaCl solution. This means that the freezing point of the \(0.1 \text{ M} \, \text{MgCl}_2\) solution is lower than that of the \(0.1 \text{ M} \, \text{NaCl}\) solution.

Conclusion

The reason the \(0.1 \text{ M} \, \text{NaCl}\) solution has a higher freezing point than the \(0.1 \text{ M} \, \text{MgCl}_2\) solution is that MgCl\(_2\) dissociates into more particles in solution (i.e., has a higher van’t Hoff factor) than NaCl, leading to a greater depression of the freezing point.

Thus, the correct answer is: 3. van’t Hoff factor for \( \text{MgCl}_2 \) is more than \( \text{NaCl} \)