The vapour pressure of pure water at `26^@C` is 25.5 torr. . The vapour pressure of a solution which contains 20.0 glucose, `(C_6H_12O_6)` , in 100 g water (in torr) is ?

Answer

To solve the problem of finding the vapor pressure of a solution containing 20.0 g of glucose (C₆H₁₂O₆) in 100 g of water at 26°C, we will use the concept of colligative properties, specifically the lowering of vapor pressure.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Vapor pressure of pure water (P₀) = 25.5 torr
- Mass of glucose (solute) = 20.0 g
- Mass of water (solvent) = 100 g

2. **Calculate Molar Mass of Glucose:**
- The molecular formula of glucose is C₆H₁₂O₆.
- Molar mass = (6 × 12) + (12 × 1) + (6 × 16) = 72 + 12 + 96 = 180 g/mol.

3. **Calculate Moles of Glucose (n₁):**
- Moles of glucose = mass / molar mass = 20 g / 180 g/mol = 0.1111 mol.

4. **Calculate Moles of Water (n₂):**
- Molar mass of water (H₂O) = (2 × 1) + (16) = 18 g/mol.
- Moles of water = mass / molar mass = 100 g / 18 g/mol = 5.5556 mol.

5. **Calculate Total Moles in Solution:**
- Total moles = moles of glucose + moles of water = 0.1111 mol + 5.5556 mol = 5.6667 mol.

6. **Calculate Mole Fraction of Solvent (X₂):**
- Mole fraction of water (X₂) = moles of water / total moles = 5.5556 mol / 5.6667 mol = 0.9804.

7. **Calculate Vapor Pressure of the Solution (Pₛ):**
- According to Raoult's Law, the vapor pressure of the solution (Pₛ) can be calculated as:
\[ Pₛ = P₀ \times X₂ \]
- Substitute the known values:
\[ Pₛ = 25.5 \, \text{torr} \times 0.9804 \approx 25.0 \, \text{torr}. \]

### Final Answer:
The vapor pressure of the solution is approximately **25.0 torr**.

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