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 will use the concept of freezing point depression, which is a colligative property. The formula for freezing point depression is:

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

Where:
- \(\Delta T_f\) = depression in freezing point
- \(i\) = van 't Hoff factor (number of particles the solute breaks into)
- \(K_f\) = freezing point depression constant (depends on the solvent, typically water)
- \(m\) = molality of the solution (moles of solute per kilogram of solvent)

### Step 1: Identify the Solutions

Let's assume we have the following solutions to analyze:
1. 1.5 m glucose (non-electrolyte)
2. 0.3 m NaCl (electrolyte)
3. 1.0 m Na2SO4 (electrolyte)
4. Pure water (H2O)

### Step 2: Calculate the van 't Hoff Factor (i)

- For glucose (C6H12O6), which is a non-electrolyte, \(i = 1\).
- For NaCl, which dissociates into Na\(^+\) and Cl\(^-\), \(i = 2\).
- For Na2SO4, which dissociates into 2 Na\(^+\) and SO4\(^{2-}\), \(i = 3\).

### Step 3: Calculate the Effective Concentration (i × m)

Now we will calculate \(i \cdot m\) for each solution:

1. **Glucose**:
\[ i \cdot m = 1 \cdot 1.5 = 1.5 \]

2. **NaCl**:
\[ i \cdot m = 2 \cdot 0.3 = 0.6 \]

3. **Na2SO4**:
\[ i \cdot m = 3 \cdot 1.0 = 3.0 \]

4. **Pure water**:
\[ i \cdot m = 0 \quad (\text{since it has no solute}) \]

### Step 4: Compare the Values

Now we compare the values of \(i \cdot m\):
- Glucose: 1.5
- NaCl: 0.6
- Na2SO4: 3.0
- Pure water: 0

### Step 5: Determine the Lowest Freezing Point

The solution with the highest value of \(i \cdot m\) will have the lowest freezing point due to a greater depression in freezing point.

From our calculations:
- The highest value is for Na2SO4 with \(i \cdot m = 3.0\).

### Conclusion

Thus, the aqueous solution that will have the lowest freezing point is the **1.0 m Na2SO4 solution**.

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