An aqueous solution boils at `101^(@)C`. What is the freezing point of the same solution?
(Gives : `K_(f) = 1.86^(@)C// m "and" K_(b) = 0.51^(@)C//m`)

Options

A. (a) `3.647^(@)C`
B. (b) `-3.647^(@)C`
C. (c) `-0.199^(@)C`
D. (d) none of these

Answer

To find the freezing point of an aqueous solution that boils at 101°C, we can follow these steps:

### Step 1: Determine the boiling point elevation (ΔTb)
The boiling point of pure water is 100°C. Since the solution boils at 101°C, we can calculate the boiling point elevation (ΔTb):
\[ \Delta T_b = T_b - T_{b, \text{pure}} = 101°C - 100°C = 1°C \]

### Step 2: Use the boiling point elevation formula
The formula for boiling point elevation is given by:
\[ \Delta T_b = i \cdot K_b \cdot m \]
Where:
- \(i\) = van 't Hoff factor (for non-electrolytes, \(i = 1\))
- \(K_b\) = molal boiling point elevation constant (given as 0.51°C/m)
- \(m\) = molality of the solution

### Step 3: Rearranging the formula to find molality (m)
Since we know ΔTb and Kb, we can rearrange the formula to find molality (m):
\[ m = \frac{\Delta T_b}{i \cdot K_b} \]
Substituting the known values:
\[ m = \frac{1°C}{1 \cdot 0.51°C/m} \approx 1.96 \, m \]

### Step 4: Use the freezing point depression formula
The formula for freezing point depression (ΔTf) is given by:
\[ \Delta T_f = -i \cdot K_f \cdot m \]
Where:
- \(K_f\) = molal freezing point depression constant (given as 1.86°C/m)

### Step 5: Calculate the freezing point depression (ΔTf)
Substituting the values we have:
\[ \Delta T_f = -1 \cdot 1.86°C/m \cdot 1.96 \, m \approx -3.647°C \]

### Step 6: Determine the freezing point of the solution
The freezing point of pure water is 0°C. Therefore, the freezing point of the solution can be calculated as:
\[ T_f = T_{f, \text{pure}} + \Delta T_f = 0°C - 3.647°C \approx -3.647°C \]

### Final Answer
The freezing point of the solution is approximately \(-3.647°C\).

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