What weight of a non-volatile solute (molar mass 40 g) should be dissolved in 114 g of octance to reduce its vapour pressure by 20%.

Answer

To solve the problem of determining the weight of a non-volatile solute needed to reduce the vapor pressure of octane by 20%, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Molar Mass of Octane**:
- The molecular formula of octane is \( C_8H_{18} \).
- Calculate the molar mass:
\[ \text{Molar mass of octane} = (8 \times 12) + (18 \times 1) = 96 + 18 = 114 \, \text{g/mol} \]

2. **Calculate the Number of Moles of Octane**:
- Given mass of octane = 114 g.
- Number of moles of octane (\( n_{octane} \)):
\[ n_{octane} = \frac{\text{mass}}{\text{molar mass}} = \frac{114 \, \text{g}}{114 \, \text{g/mol}} = 1 \, \text{mol} \]

3. **Define the Weight of the Solute**:
- Let the weight of the non-volatile solute be \( x \) grams.
- The molar mass of the solute is given as 40 g/mol.
- Number of moles of solute (\( n_{solute} \)):
\[ n_{solute} = \frac{x}{40} \]

4. **Calculate Total Number of Moles**:
- Total number of moles in the solution:
\[ n_{total} = n_{octane} + n_{solute} = 1 + \frac{x}{40} \]

5. **Determine the Change in Vapor Pressure**:
- The problem states that the vapor pressure is reduced by 20%. This means the new vapor pressure (\( P_2 \)) is:
\[ P_2 = P_0 - 0.2P_0 = 0.8P_0 \]
- The lowering of vapor pressure (\( \Delta P \)):
\[ \Delta P = P_0 - P_2 = 0.2P_0 \]

6. **Relate Lowering of Vapor Pressure to Mole Fraction**:
- According to Raoult’s Law:
\[ \frac{\Delta P}{P_0} = \text{mole fraction of solute} \]
- Thus:
\[ \frac{0.2P_0}{P_0} = \frac{n_{solute}}{n_{total}} \]
- This simplifies to:
\[ 0.2 = \frac{\frac{x}{40}}{1 + \frac{x}{40}} \]

7. **Cross-Multiply and Solve for \( x \)**:
- Cross-multiplying gives:
\[ 0.2 \left(1 + \frac{x}{40}\right) = \frac{x}{40} \]
- Expanding this:
\[ 0.2 + \frac{0.2x}{40} = \frac{x}{40} \]
- Rearranging terms:
\[ 0.2 = \frac{x}{40} - \frac{0.2x}{40} \]
\[ 0.2 = \frac{0.8x}{40} \]
- Multiplying both sides by 40:
\[ 8 = 0.8x \]
- Solving for \( x \):
\[ x = \frac{8}{0.8} = 10 \, \text{g} \]

### Final Answer:
The weight of the non-volatile solute that should be dissolved in 114 g of octane to reduce its vapor pressure by 20% is **10 grams**.