The freezing point of 0.1 M solution of glucose is `-1.86^(@)C`. If an equal volume of 0.3 M glucose solution is added, the freezing point of the mixture will be

Options

A. `-7.44^(@)C`
B. `-5.58^(@)C`
C. `-3.72^(@)C`
D. `-2.79^(@)`

Explanation

To solve the problem of finding the freezing point of the mixture after combining a 0.1 M glucose solution with a 0.3 M glucose solution, we can follow these steps:

### Step 1: Understand the Depression in Freezing Point
The depression in freezing point (ΔTf) is given by the formula:
\[ \Delta T_f = K_f \cdot m \cdot i \]
where:
- \( K_f \) is the freezing point depression constant,
- \( m \) is the molality of the solution,
- \( i \) is the van 't Hoff factor (which is 1 for glucose since it does not dissociate).

### Step 2: Calculate the Freezing Point Depression of the 0.1 M Solution
From the problem, we know that the freezing point of a 0.1 M glucose solution is -1.86 °C. The depression in freezing point can be calculated as:
\[ \Delta T_f = T_{\text{solvent}} - T_{\text{solution}} = 0 - (-1.86) = 1.86 \, °C \]

### Step 3: Calculate the Freezing Point Depression Constant (Kf)
Using the depression in freezing point calculated above, we can find \( K_f \):
\[ \Delta T_f = K_f \cdot m \cdot i \]
Substituting \( \Delta T_f = 1.86 \), \( m = 0.1 \), and \( i = 1 \):
\[ 1.86 = K_f \cdot 0.1 \cdot 1 \]
\[ K_f = \frac{1.86}{0.1} = 18.6 \, °C \, \text{per molal} \]

### Step 4: Calculate the Molarity of the Mixture
Next, we need to find the molarity of the new mixture when equal volumes of 0.1 M and 0.3 M glucose solutions are combined. The total volume will be \( 2V \) (where \( V \) is the volume of each solution).

Using the formula for mixing solutions:
\[ M_{\text{mixture}} = \frac{M_1V_1 + M_2V_2}{V_1 + V_2} \]
Substituting the values:
\[ M_{\text{mixture}} = \frac{(0.1 \, \text{M} \cdot V) + (0.3 \, \text{M} \cdot V)}{V + V} \]
\[ = \frac{(0.1 + 0.3)V}{2V} = \frac{0.4}{2} = 0.2 \, \text{M} \]

### Step 5: Calculate the Depression in Freezing Point for the Mixture
Now we can calculate the depression in freezing point for the 0.2 M glucose solution:
\[ \Delta T_f = K_f \cdot m \cdot i \]
Substituting \( K_f = 18.6 \, °C \, \text{per molal} \), \( m = 0.2 \), and \( i = 1 \):
\[ \Delta T_f = 18.6 \cdot 0.2 \cdot 1 = 3.72 \, °C \]

### Step 6: Determine the Freezing Point of the Mixture
Finally, we can find the freezing point of the mixture:
\[ T_f = T_{\text{solvent}} - \Delta T_f = 0 - 3.72 = -3.72 \, °C \]

### Conclusion
The freezing point of the mixture is:
\[ \boxed{-3.72 \, °C} \]