Two liquids, `A` and`B` form an ideal solution. At the specified temperature, the vapour pressure of pure `A` is `20 mm Hg` while that of pure `B` is `75 mm Hg`. If the vapour over the mixture consists of `50` mol percent `A`, what is the mole percent `A` in the liquid?

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the given data
We have two liquids, A and B, that form an ideal solution. The vapor pressures of the pure components are:
- Vapor pressure of pure A (P°A) = 20 mm Hg
- Vapor pressure of pure B (P°B) = 75 mm Hg

The vapor over the mixture consists of 50 mol% A, which means:
- Mole fraction of A in vapor (Y_A) = 0.5
- Mole fraction of B in vapor (Y_B) = 1 - Y_A = 0.5

### Step 2: Use Raoult's Law
According to Raoult's Law, the partial pressures of the components in the vapor phase can be expressed as:
- Partial pressure of A (P_A) = P°A * X_A
- Partial pressure of B (P_B) = P°B * X_B

Where:
- X_A = mole fraction of A in the liquid
- X_B = mole fraction of B in the liquid

### Step 3: Relate the mole fractions
From the definition of mole fractions in the vapor phase, we have:
- Y_A = P_A / (P_A + P_B)
- Y_B = P_B / (P_A + P_B)

Since Y_A = 0.5 and Y_B = 0.5, we can equate the partial pressures:
- P_A = P_B

### Step 4: Substitute the expressions for partial pressures
Using the expressions from Raoult's Law:
- P°A * X_A = P°B * X_B

### Step 5: Express X_B in terms of X_A
Since X_B = 1 - X_A, we can substitute:
- P°A * X_A = P°B * (1 - X_A)

### Step 6: Substitute the known vapor pressures
Substituting the values of P°A and P°B:
- 20 * X_A = 75 * (1 - X_A)

### Step 7: Solve for X_A
Expanding and rearranging the equation:
- 20 * X_A = 75 - 75 * X_A
- 20 * X_A + 75 * X_A = 75
- 95 * X_A = 75
- X_A = 75 / 95
- X_A = 0.7895 (approximately)

### Step 8: Convert mole fraction to mole percent
To find the mole percent of A in the liquid:
- Mole percent of A = X_A * 100 = 0.7895 * 100 = 78.95%

### Final Answer
The mole percent of A in the liquid is approximately **79%**.

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