Estimste the lowering of vapour pressure due to the solute (glucose) in a 1.0 M aqueous solution at `100^(@)C`:

Options

A. 10 torr
B. 18 torr
C. 13.45 torr
D. 24 torr

Answer

To estimate the lowering of vapor pressure due to the solute (glucose) in a 1.0 M aqueous solution at 100°C, we can follow these steps:

### Step 1: Understand the Given Information
We have a 1.0 M solution of glucose in water at 100°C. We need to find the lowering of vapor pressure due to the presence of glucose.

### Step 2: Identify the Relevant Concepts
The lowering of vapor pressure is a colligative property, which depends on the number of solute particles in a solution and not on their identity. We will use Raoult's Law for this calculation.

### Step 3: Calculate the Moles of Solute and Solvent
1. **Moles of Solute (Glucose)**:
- Given that the solution is 1.0 M, this means there is 1 mole of glucose in 1 liter of solution.

2. **Moles of Solvent (Water)**:
- The density of water is approximately 1 g/mL, so 1 liter of water has a mass of 1000 g.
- The molar mass of water (H₂O) is approximately 18 g/mol.
- Moles of water = mass of water / molar mass of water = 1000 g / 18 g/mol ≈ 55.56 moles.

### Step 4: Calculate the Mole Fraction of the Solute
The mole fraction of the solute (glucose) can be calculated using the formula:
\[ x_2 = \frac{n_2}{n_1 + n_2} \]
Where:
- \( n_2 \) = moles of solute (glucose) = 1 mole
- \( n_1 \) = moles of solvent (water) ≈ 55.56 moles

\[ x_2 = \frac{1}{55.56 + 1} \approx \frac{1}{56.56} \approx 0.0177 \]

### Step 5: Apply Raoult's Law
According to Raoult's Law, the relative lowering of vapor pressure is given by:
\[ \frac{P_0 - P_s}{P_0} = x_2 \]
Where:
- \( P_0 \) is the vapor pressure of pure solvent (water at 100°C ≈ 760 mmHg).
- \( P_s \) is the vapor pressure of the solution.

Rearranging gives:
\[ P_s = P_0 (1 - x_2) \]

### Step 6: Calculate the Vapor Pressure of the Solution
Substituting the values:
\[ P_s = 760 \times (1 - 0.0177) \approx 760 \times 0.9823 \approx 746.77 \text{ mmHg} \]

### Step 7: Calculate the Lowering of Vapor Pressure
The lowering of vapor pressure is:
\[ P_0 - P_s = 760 - 746.77 \approx 13.23 \text{ mmHg} \]

### Step 8: Final Result
The estimated lowering of vapor pressure due to glucose in a 1.0 M aqueous solution at 100°C is approximately **13.23 mmHg**.

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