The vapour pressure of a pure solvent at a certain temperature is 0.0227 bar. What is the vapour pressure of a solution containing 6 g of solute (M=60 g/mol) in 50 g of solvent?

Answer

To find the vapor pressure of a solution containing 6 g of solute in 50 g of solvent, we can use Raoult's Law, which states that the vapor pressure of a solution is directly proportional to the mole fraction of the solvent in the solution.

### Step-by-Step Solution:

**Step 1: Calculate the number of moles of solute.**
- Given mass of solute = 6 g
- Molar mass of solute (M) = 60 g/mol

\[ \text{Number of moles of solute} = \frac{\text{mass of solute}}{\text{molar mass of solute}} = \frac{6 \, \text{g}}{60 \, \text{g/mol}} = 0.1 \, \text{mol} \]

**Step 2: Calculate the number of moles of solvent.**
- Given mass of solvent = 50 g
- Molar mass of solvent (assuming it is water) = 18 g/mol

\[ \text{Number of moles of solvent} = \frac{\text{mass of solvent}}{\text{molar mass of solvent}} = \frac{50 \, \text{g}}{18 \, \text{g/mol}} \approx 2.78 \, \text{mol} \]

**Step 3: Calculate the total number of moles in the solution.**
\[ \text{Total moles} = \text{moles of solute} + \text{moles of solvent} = 0.1 \, \text{mol} + 2.78 \, \text{mol} \approx 2.88 \, \text{mol} \]

**Step 4: Calculate the mole fraction of the solvent.**
\[ \text{Mole fraction of solvent} (X_{solvent}) = \frac{\text{moles of solvent}}{\text{total moles}} = \frac{2.78 \, \text{mol}}{2.88 \, \text{mol}} \approx 0.967 \]

**Step 5: Calculate the vapor pressure of the solution using Raoult's Law.**
- Vapor pressure of pure solvent (P°) = 0.0227 bar

\[ P_{solution} = X_{solvent} \times P° = 0.967 \times 0.0227 \, \text{bar} \approx 0.0219 \, \text{bar} \]

### Final Answer:
The vapor pressure of the solution is approximately **0.0219 bar**.