What will be the molecular weight of `NaCl` determined experimentally following elevation in the boiling point or depression in freezing point method?

Options

A. `lt58.5`
B. `gt58.5`
C. `=58.5`
D. None of these

Explanation

To determine the molecular weight of NaCl experimentally using the elevation in boiling point or depression in freezing point method, we can follow these steps:

### Step-by-Step Solution:

1. **Understand Colligative Properties**:
- Colligative properties depend on the number of solute particles in a solution, not on the identity of the solute. The two main colligative properties we will consider are boiling point elevation and freezing point depression.

2. **Identify the Van't Hoff Factor (i)**:
- For NaCl, which dissociates into two ions (Na⁺ and Cl⁻) when it dissolves in water, the Van't Hoff factor (i) is 2. This means that one formula unit of NaCl produces two particles in solution.

3. **Use the Formula for Colligative Properties**:
- The change in boiling point (ΔT_b) or the change in freezing point (ΔT_f) can be related to the molality (m) of the solution and the Van't Hoff factor (i) using the formulas:
\[ \Delta T_b = i \cdot K_b \cdot m \]
\[ \Delta T_f = i \cdot K_f \cdot m \]
- Here, \(K_b\) and \(K_f\) are the ebullioscopic and cryoscopic constants of the solvent (water in this case).

4. **Calculate the Experimental Molecular Weight**:
- Rearranging the formulas, we can express the molality in terms of the change in temperature and the constants:
\[ m = \frac{\Delta T_b}{i \cdot K_b} \quad \text{or} \quad m = \frac{\Delta T_f}{i \cdot K_f} \]
- The molality (m) can also be related to the number of moles of solute and the mass of the solvent:
\[ m = \frac{n}{\text{mass of solvent (kg)}} \]
- Where \(n\) is the number of moles of solute.

5. **Determine the Molar Mass**:
- The molar mass (M) of NaCl can be calculated using the relationship:
\[ M = \frac{\text{mass of solute (g)}}{n} \]
- Since \(n\) can be expressed in terms of molality and the mass of the solvent, we can substitute to find the molar mass.

6. **Conclusion**:
- Given that the calculated molar mass of NaCl is approximately 58.5 g/mol, and considering the dissociation factor (i = 2), the experimentally determined molar mass will be higher due to the effect of colligative properties. Thus, the experimental molar mass of NaCl will be greater than 58.5 g/mol.

### Final Answer:
The molecular weight of NaCl determined experimentally following elevation in the boiling point or depression in freezing point method is greater than 58.5 g/mol.