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^(@)C`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the freezing point depression formula
The freezing point depression (\(\Delta T_f\)) can be calculated using the formula:
\[ \Delta T_f = k_f \cdot m \]
where:
- \(k_f\) is the freezing point depression constant,
- \(m\) is the molality of the solution.

### Step 2: Calculate \(k_f\) using the first solution
We know that the freezing point of a 0.1 M glucose solution is -1.86°C. The freezing point of pure water is 0°C. Thus, we can calculate \(\Delta T_f\) for the 0.1 M solution:
\[ \Delta T_f = 0 - (-1.86) = 1.86°C \]

Now, we can find \(k_f\):
\[ k_f = \frac{\Delta T_f}{m} = \frac{1.86}{0.1} = 18.6 \, \text{°C kg/mol} \]

### Step 3: Calculate the new molality after mixing
We have two solutions being mixed:
- 0.1 M glucose solution (volume = \(V\))
- 0.3 M glucose solution (volume = \(V\))

When mixed, the total volume becomes \(2V\). The total moles of glucose in the mixture is:
\[ \text{Moles from 0.1 M solution} = 0.1 \cdot V \]
\[ \text{Moles from 0.3 M solution} = 0.3 \cdot V \]
Total moles of glucose:
\[ \text{Total moles} = 0.1V + 0.3V = 0.4V \]

The molarity of the resulting solution can be calculated as:
\[ \text{Molarity} = \frac{\text{Total moles}}{\text{Total volume}} = \frac{0.4V}{2V} = 0.2 \, \text{M} \]

### Step 4: Calculate the freezing point depression for the new solution
Now we can use the new molality (which is equal to molarity for dilute solutions) to find the new freezing point depression:
\[ \Delta T_f = k_f \cdot m = 18.6 \cdot 0.2 = 3.72 \, \text{°C} \]

### Step 5: Determine the new freezing point of the mixture
The freezing point of the mixture will be:
\[ \text{Freezing point} = 0 - \Delta T_f = 0 - 3.72 = -3.72 \, \text{°C} \]

### Final Answer
The freezing point of the mixture will be \(-3.72 \, \text{°C}\).

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