The freezing point of a 0.01 M aqueous glucose solution at 1 atmosphere is `-0.18^(@)C`. To it, an addition of equal volume of 0.002 M glucose solution will , produced a solution with freezing point of nearly

Options

A. `-0.036^(@)C`
B. `-0.108^(@)C`
C. `-0.216^(@)C`
D. `-0.422^(@)C`

Explanation

To solve the problem step-by-step, we will follow these steps:

### Step 1: Understand the Given Data
We have a 0.01 M aqueous glucose solution that has a freezing point of -0.18°C. We are adding an equal volume of a 0.002 M glucose solution to this.

### Step 2: Calculate the Resultant Molarity
When two solutions of equal volume are mixed, the resultant molarity can be calculated using the formula:

\[ M_{\text{resultant}} = \frac{M_1V_1 + M_2V_2}{V_1 + V_2} \]

Since the volumes are equal, we can denote \(V_1 = V_2 = V\). Thus, the formula simplifies to:

\[ M_{\text{resultant}} = \frac{M_1 + M_2}{2} \]

Substituting the values:

\[ M_{\text{resultant}} = \frac{0.01 + 0.002}{2} = \frac{0.012}{2} = 0.006 \, \text{M} \]

### Step 3: Use 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 cryoscopic constant (which remains constant for a given solvent),
- \(m\) is the molality of the solution.

For dilute solutions, molality and molarity can be approximated to be equal, so we can use the molarity directly.

### Step 4: Calculate the Depression in Freezing Point
We know the depression in freezing point for the first solution is:

\[ \Delta T_{f1} = 0.18 \, \text{°C} \quad \text{(for 0.01 M)} \]

Now, we can set up a ratio to find the depression for the resultant solution:

\[ \frac{\Delta T_{f1}}{\Delta T_{f2}} = \frac{M_1}{M_2} \]

Where:
- \(M_1 = 0.01 \, \text{M}\) (initial solution),
- \(M_2 = 0.006 \, \text{M}\) (resultant solution).

Substituting the values:

\[ \frac{0.18}{\Delta T_{f2}} = \frac{0.01}{0.006} \]

### Step 5: Solve for \(\Delta T_{f2}\)
Cross-multiplying gives:

\[ 0.18 \cdot 0.006 = 0.01 \cdot \Delta T_{f2} \]

Calculating the left side:

\[ 0.00108 = 0.01 \cdot \Delta T_{f2} \]

Now, solving for \(\Delta T_{f2}\):

\[ \Delta T_{f2} = \frac{0.00108}{0.01} = 0.108 \, \text{°C} \]

### Step 6: Calculate the Final Freezing Point
The final freezing point of the solution is given by:

\[ T_f = T_{f0} - \Delta T_f \]

Where \(T_{f0} = 0 \, \text{°C}\) (freezing point of pure water):

\[ T_f = 0 - 0.108 = -0.108 \, \text{°C} \]

### Final Answer
The freezing point of the resultant solution is approximately **-0.108°C**.

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