The vapour pressure of a pure liquid A at 300 K is 150 torr. The vapour pressure of this liquid in a solution with a liquid Bis 105 torr. Calculate the mole fraction of A in the solution if the mixture obey's Raoult's law.

Explanation

To solve the problem, we will use Raoult's Law, which states that the vapor pressure of a solvent in a solution is directly proportional to the mole fraction of the solvent in the solution.

### Given Data:
- Vapor pressure of pure liquid A (P°_A) = 150 torr
- Vapor pressure of liquid A in the solution (P_A) = 105 torr

### Step 1: Apply Raoult's Law
According to Raoult's Law, the vapor pressure of component A in the solution is given by:
\[ P_A = \chi_A \cdot P°_A \]
where:
- \( P_A \) = vapor pressure of A in the solution
- \( \chi_A \) = mole fraction of A in the solution
- \( P°_A \) = vapor pressure of pure A

### Step 2: Rearrange the equation to find the mole fraction of A
We can rearrange the equation to solve for the mole fraction of A:
\[ \chi_A = \frac{P_A}{P°_A} \]

### Step 3: Substitute the known values
Now, substitute the given values into the equation:
\[ \chi_A = \frac{105 \text{ torr}}{150 \text{ torr}} \]

### Step 4: Calculate the mole fraction
Now we perform the calculation:
\[ \chi_A = \frac{105}{150} = 0.7 \]

### Step 5: Conclusion
The mole fraction of A in the solution is 0.7.