Which solution will show the maximum vapour pressure at `300K`

Options

A. `1M NaCl`
B. `1 M AlCl_(3)`
C. `1M CaCl_(2)`
D. `1M C_(12)H_(22)O_(11)`

Explanation

To determine which solution will show the maximum vapor pressure at 300 K, we need to analyze the effect of solute particles on vapor pressure according to colligative properties. Here’s a step-by-step solution:

### Step 1: Understand Colligative Properties
Colligative properties depend on the number of solute particles in a solution, not on the nature of the solute. The key colligative property we are interested in here is vapor pressure lowering.

### Step 2: Identify the Solutions
We have four solutions to consider:
1. NaCl (1 molar)
2. AlCl3 (1 molar)
3. CaCl2 (1 molar)
4. C12H22O11 (1 molar)

### Step 3: Determine the Number of Particles Produced by Each Solute
- **NaCl** dissociates into 2 particles: Na⁺ and Cl⁻.
- **AlCl3** dissociates into 4 particles: Al³⁺ and 3 Cl⁻.
- **CaCl2** dissociates into 3 particles: Ca²⁺ and 2 Cl⁻.
- **C12H22O11** (sucrose) does not dissociate, so it remains as 1 particle.

### Step 4: Calculate the Total Number of Particles
- NaCl: 2 particles
- AlCl3: 4 particles
- CaCl2: 3 particles
- C12H22O11: 1 particle

### Step 5: Relate Particle Number to Vapor Pressure
The vapor pressure lowering is directly proportional to the number of solute particles. The more particles present, the greater the lowering of vapor pressure.

### Step 6: Identify the Solution with the Least Vapor Pressure Lowering
Since C12H22O11 has the least number of particles (1), it will cause the least lowering of vapor pressure. Thus, it will have the maximum vapor pressure among the four solutions.

### Conclusion
The solution that will show the maximum vapor pressure at 300 K is **C12H22O11 (1 molar)**.

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