A solution of glucose (Molar mass `="180 g mol"^(-1)`) in water has a boiling of `100.20^(@)C`. Calculate the freezing point of the same solution. Molal constants for water `K_(f) and K_(b)` are `"1.86 K kg mol"^(-1)` and 0.512 K kg `"mol"^(-1)` respectively.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the elevation in boiling point (ΔT_b)
The boiling point of the solution is given as 100.20°C, and the boiling point of pure water (T_b) is 100.00°C.

\[ \Delta T_b = T_b(\text{solution}) - T_b(\text{water}) = 100.20°C - 100.00°C = 0.20°C \]

### Step 2: Use the boiling point elevation formula to find molality (m)
The formula for boiling point elevation is:

\[ \Delta T_b = K_b \cdot m \]

Where:
- \( K_b = 0.512 \, \text{K kg mol}^{-1} \) (this value is the same in °C)
- \( \Delta T_b = 0.20°C \)

Rearranging the formula to find molality (m):

\[ m = \frac{\Delta T_b}{K_b} = \frac{0.20°C}{0.512 \, \text{K kg mol}^{-1}} \approx 0.3906 \, \text{mol kg}^{-1} \]

### Step 3: Calculate the depression in freezing point (ΔT_f)
The formula for freezing point depression is:

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

Where:
- \( K_f = 1.86 \, \text{K kg mol}^{-1} \)

Substituting the values:

\[ \Delta T_f = 1.86 \, \text{K kg mol}^{-1} \cdot 0.3906 \, \text{mol kg}^{-1} \approx 0.726 \, \text{K} \]

### Step 4: Calculate the freezing point of the solution (T_f)
The freezing point depression formula is:

\[ \Delta T_f = T_f^0 - T_f \]

Where:
- \( T_f^0 = 0°C \) (freezing point of pure water)

Rearranging the formula to find the freezing point of the solution (T_f):

\[ T_f = T_f^0 - \Delta T_f = 0°C - 0.726°C \approx -0.726°C \]

### Final Answer
The freezing point of the glucose solution is approximately \(-0.73°C\).

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