In coparison to a 0.01 M solution of glucose, the depression in freezing point of a 0.01 M `MgCl_(2)` solution is……

Options

A. the same
B. about twice
C. about three times
D. about six times

Explanation

To solve the problem of comparing the depression in freezing point of a 0.01 M solution of glucose with that of a 0.01 M solution of MgCl₂, we will follow these steps:

### Step 1: Understand the formula for depression in freezing point
The depression in freezing point (ΔTf) can be calculated using the formula:
\[ \Delta T_f = i \cdot K_f \cdot m \]
where:
- \( \Delta T_f \) = depression in freezing point
- \( i \) = van't Hoff factor (number of particles the solute dissociates into)
- \( K_f \) = molal freezing point depression constant (depends on the solvent)
- \( m \) = molality of the solution

### Step 2: Identify the values for glucose
For glucose (C₆H₁₂O₆):
- It does not dissociate in solution, so the van't Hoff factor \( i \) for glucose is 1.
- The molality \( m \) is given as 0.01 M.

### Step 3: Identify the values for MgCl₂
For magnesium chloride (MgCl₂):
- It dissociates into three ions in solution: one magnesium ion (Mg²⁺) and two chloride ions (Cl⁻). Therefore, the van't Hoff factor \( i \) for MgCl₂ is 3.
- The molality \( m \) is also given as 0.01 M.

### Step 4: Compare the depression in freezing points
Since the molality and the \( K_f \) constant are the same for both solutions, we can focus on the van't Hoff factors:
- For glucose: \( \Delta T_f = 1 \cdot K_f \cdot 0.01 \)
- For MgCl₂: \( \Delta T_f = 3 \cdot K_f \cdot 0.01 \)

### Step 5: Conclusion
Thus, the depression in freezing point of the MgCl₂ solution is three times that of the glucose solution:
\[ \Delta T_f \text{ (MgCl₂)} = 3 \cdot \Delta T_f \text{ (glucose)} \]

### Final Answer
In comparison to a 0.01 M solution of glucose, the depression in freezing point of a 0.01 M MgCl₂ solution is **three times greater**.

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