20 moles of liquid A are mixed with 20 moles of liquid B to form an ideal binary solution. Calculate moles of A in liquid state when half of the solution has vaporized :
( Given : `P_(A)^(@)=100 " torr", P_(B)^(@)=121 " torr"`)

Options

A. `(11)/(21)`
B. `(110)/(21)`
C. `(200)/(21)`
D. `(220)/(21)`

Answer

To solve the problem step by step, we need to calculate the number of moles of liquid A remaining in the liquid state when half of the solution has vaporized.

### Step 1: Understand the Problem
We have 20 moles of liquid A and 20 moles of liquid B mixed to form an ideal binary solution. When half of the solution vaporizes, we need to find out how many moles of liquid A remain in the liquid state.

### Step 2: Calculate the Total Initial Moles
Initially, we have:
- Moles of A = 20
- Moles of B = 20
- Total moles = 20 + 20 = 40 moles

### Step 3: Determine the Moles After Vaporization
When half of the solution vaporizes, the total moles of the solution that remain in the liquid state will be:
- Remaining moles = 40 moles / 2 = 20 moles

### Step 4: Use Raoult's Law
According to Raoult's Law, the total vapor pressure (P_total) of the solution can be calculated using the formula:
\[ P_{total} = X_A \cdot P_{A}^0 + X_B \cdot P_{B}^0 \]
Where:
- \( P_{A}^0 = 100 \, \text{torr} \) (vapor pressure of pure A)
- \( P_{B}^0 = 121 \, \text{torr} \) (vapor pressure of pure B)

### Step 5: Calculate the Total Vapor Pressure
The total vapor pressure when half the solution has vaporized can be calculated as:
\[ P_{total} = \sqrt{P_{A}^0 \cdot P_{B}^0} = \sqrt{100 \cdot 121} = \sqrt{12100} = 110 \, \text{torr} \]

### Step 6: Set Up the Mole Fraction Equation
Let \( X_A \) be the mole fraction of A in the liquid phase and \( X_B \) be the mole fraction of B in the liquid phase. Since we have 20 moles remaining in total:
- \( X_A = \frac{n_A}{20} \)
- \( X_B = \frac{n_B}{20} \)

Where \( n_A \) and \( n_B \) are the moles of A and B remaining in the liquid state respectively.

### Step 7: Express the Mole Fractions
From the total moles remaining:
- \( n_A + n_B = 20 \)
- \( n_B = 20 - n_A \)

### Step 8: Substitute into Raoult's Law
Substituting the mole fractions into Raoult's Law:
\[ 110 = \left(\frac{n_A}{20}\right) \cdot 100 + \left(\frac{20 - n_A}{20}\right) \cdot 121 \]

### Step 9: Simplify the Equation
Multiply through by 20 to eliminate the denominator:
\[ 2200 = 100n_A + 121(20 - n_A) \]
\[ 2200 = 100n_A + 2420 - 121n_A \]
\[ 2200 = 2420 - 21n_A \]

### Step 10: Solve for \( n_A \)
Rearranging gives:
\[ 21n_A = 2420 - 2200 \]
\[ 21n_A = 220 \]
\[ n_A = \frac{220}{21} \]

### Final Answer
The number of moles of liquid A remaining in the liquid state when half of the solution has vaporized is:
\[ n_A \approx 10.48 \, \text{moles} \]