Lowering in vapour pressure is highest for

Options

A. `0.2 m urea`
B. `0.1 m glucose`
C. `0.1 m MgSO_(4)`
D. `0.1 m BaCl_(2)`

Explanation

To determine which solution has the highest lowering in vapor pressure, we can follow these steps:

### Step 1: Understand the Concept of Lowering in Vapor Pressure
Lowering in vapor pressure occurs when a non-volatile solute is added to a solvent. The relative lowering in vapor pressure can be expressed as:
\[ \frac{P^0 - P_s}{P^0} = \chi_B \]
where \( P^0 \) is the vapor pressure of the pure solvent, \( P_s \) is the vapor pressure of the solution, and \( \chi_B \) is the mole fraction of the solute.

### Step 2: Consider the Van 't Hoff Factor
For ionic compounds, we need to consider the Van 't Hoff factor (i), which accounts for the number of particles the solute dissociates into. The formula becomes:
\[ \frac{P^0 - P_s}{P^0} = \frac{n_B}{n_A} \cdot i \]
where \( n_B \) is the number of moles of solute and \( n_A \) is the number of moles of solvent.

### Step 3: Calculate the Effective Concentration for Each Solute
We will calculate \( M \cdot i \) for each of the given solutes to find which has the highest lowering in vapor pressure.

1. **Urea (0.2 molal, i = 1)**:
\[ M \cdot i = 0.2 \cdot 1 = 0.2 \]

2. **Glucose (0.1 molal, i = 1)**:
\[ M \cdot i = 0.1 \cdot 1 = 0.1 \]

3. **MgSO₄ (0.1 molal, i = 2)**:
\[ M \cdot i = 0.1 \cdot 2 = 0.2 \]

4. **BaCl₂ (0.1 molal, i = 3)**:
\[ M \cdot i = 0.1 \cdot 3 = 0.3 \]

### Step 4: Compare the Results
Now we compare the values of \( M \cdot i \):
- Urea: 0.2
- Glucose: 0.1
- MgSO₄: 0.2
- BaCl₂: 0.3

### Step 5: Conclusion
The solution with the highest lowering in vapor pressure is the one with BaCl₂, as it has the highest value of \( M \cdot i = 0.3 \).

### Final Answer
**BaCl₂ has the highest lowering in vapor pressure.**

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