`100g` of liquid `A(` molar mass `140 g mol ^(-1))` was dissolved in `1000g ` of liquid `B(` molar mass `180g mol^(-1))`. The vapour pressure of pure liquid `B` was found to be `500` torr. Calculate the vapour pressure of pure liquid `A` and its vapour pressure in the solution if the total vapour pressure of the solution is `475 Tor r`

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Calculate the number of moles of liquid B
Given:
- Mass of liquid B = 1000 g
- Molar mass of liquid B = 180 g/mol

Using the formula for moles:
\[ \text{Number of moles of B} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{1000 \, \text{g}}{180 \, \text{g/mol}} = 5.56 \, \text{mol} \]

### Step 2: Calculate the number of moles of liquid A
Given:
- Mass of liquid A = 100 g
- Molar mass of liquid A = 140 g/mol

Using the formula for moles:
\[ \text{Number of moles of A} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{100 \, \text{g}}{140 \, \text{g/mol}} = 0.714 \, \text{mol} \]

### Step 3: Calculate the mole fraction of liquid B
The total number of moles in the solution is:
\[ \text{Total moles} = \text{Moles of A} + \text{Moles of B} = 0.714 + 5.56 = 6.274 \, \text{mol} \]

Now, calculate the mole fraction of B (\(X_B\)):
\[ X_B = \frac{\text{Moles of B}}{\text{Total moles}} = \frac{5.56}{6.274} = 0.886 \]

### Step 4: Calculate the mole fraction of liquid A
Using the relationship \(X_A + X_B = 1\):
\[ X_A = 1 - X_B = 1 - 0.886 = 0.114 \]

### Step 5: Calculate the vapor pressure of liquid B in the solution
Using Raoult's Law:
\[ P_B = P_{0B} \times X_B \]
Given \(P_{0B} = 500 \, \text{torr}\):
\[ P_B = 500 \, \text{torr} \times 0.886 = 443 \, \text{torr} \]

### Step 6: Use the total vapor pressure to find the vapor pressure of liquid A in the solution
Given the total vapor pressure \(P_{\text{total}} = 475 \, \text{torr}\):
\[ P_{\text{total}} = P_A + P_B \]
Rearranging gives:
\[ P_A = P_{\text{total}} - P_B = 475 \, \text{torr} - 443 \, \text{torr} = 32 \, \text{torr} \]

### Step 7: Calculate the vapor pressure of pure liquid A
Using Raoult's Law again:
\[ P_A = P_{0A} \times X_A \]
Rearranging gives:
\[ P_{0A} = \frac{P_A}{X_A} = \frac{32 \, \text{torr}}{0.114} \approx 280.7 \, \text{torr} \]

### Summary of Results
- Vapor pressure of pure liquid A (\(P_{0A}\)): **280.7 torr**
- Vapor pressure of liquid A in the solution (\(P_A\)): **32 torr**