Which of the following aqueous solutions will have the lowest freezing point ?

Options

A. 1.5 M glucose
B. 0.3 M `Na_2SO_4`
C. 1 M NaCl
D. `H_2O`

Explanation

To determine which aqueous solution has the lowest freezing point, we can use the formula for freezing point depression:

\[ \Delta T_f = i \cdot K_f \cdot m \]

Where:
- \(\Delta T_f\) = depression in freezing point
- \(i\) = Van't Hoff factor (number of particles the solute dissociates into)
- \(K_f\) = molal freezing point depression constant (a constant for the solvent)
- \(m\) = molality of the solution

Since \(K_f\) is constant for water, we can focus on maximizing the product \(i \cdot m\) to find which solution will have the lowest freezing point.

### Step-by-Step Solution:

1. **Identify the Van't Hoff factor (i) for each solute:**
- **Glucose (C6H12O6)**: Does not dissociate in solution, so \(i = 1\).
- **Na2SO4**: Dissociates into 2 Na⁺ ions and 1 SO₄²⁻ ion, so \(i = 3\).
- **NaCl**: Dissociates into 1 Na⁺ ion and 1 Cl⁻ ion, so \(i = 2\).
- **H2O**: Pure water has no solute, so it is not applicable for this question.

2. **Calculate the product \(i \cdot m\) for each solution:**
- **1.5 M Glucose**:
\[ i \cdot m = 1 \cdot 1.5 = 1.5 \]
- **0.3 M Na2SO4**:
\[ i \cdot m = 3 \cdot 0.3 = 0.9 \]
- **1 M NaCl**:
\[ i \cdot m = 2 \cdot 1 = 2 \]

3. **Compare the values of \(i \cdot m\)**:
- Glucose: 1.5
- Na2SO4: 0.9
- NaCl: 2

4. **Determine which solution has the highest \(i \cdot m\)**:
- The highest value is for **1 M NaCl**, which has a product of \(2\).

5. **Conclusion**:
- The solution with the lowest freezing point is **1 M NaCl** because it has the highest product of \(i \cdot m\), which leads to the greatest depression in freezing point.

### Final Answer:
**1 M NaCl will have the lowest freezing point.**