To solve the problem of finding the mass of the immiscible liquid present per gram of water in the distillate, we can follow these steps:
### Step 1: Gather Given Data
- Molar mass of the liquid (A), \( M_A = 134.3 \, \text{g/mol} \)
- Vapor pressure of the liquid (A), \( P^0_A = 0.143 \, \text{atm} \)
- Vapor pressure of water (B), \( P^0_B = 0.84 \, \text{atm} \)
- Total pressure during distillation, \( P_T = 0.983 \, \text{atm} \)
- Mass of water used, \( W_B = 1 \, \text{g} \)
- Molar mass of water, \( M_B = 18 \, \text{g/mol} \)
### Step 2: Use Dalton's Law of Partial Pressures
According to Dalton's law, the partial pressure of each component in a mixture is given by:
\[
P_A = X_A \cdot P_T
\]
\[
P_B = X_B \cdot P_T
\]
Where \( X_A \) and \( X_B \) are the mole fractions of the liquid and water, respectively.
### Step 3: Calculate Mole Fractions
The total pressure \( P_T \) is the sum of the partial pressures:
\[
P_T = P_A + P_B
\]
Using the vapor pressures:
\[
P_A = P^0_A \cdot X_A
\]
\[
P_B = P^0_B \cdot X_B
\]
### Step 4: Set Up the Equations
From the above equations, we can express the mole fractions:
\[
X_A = \frac{P_A}{P_T} \quad \text{and} \quad X_B = \frac{P_B}{P_T}
\]
Since \( X_A + X_B = 1 \), we can substitute \( X_B \) as \( 1 - X_A \).
### Step 5: Substitute Values
Using the known vapor pressures:
\[
P_T = P^0_A \cdot X_A + P^0_B \cdot (1 - X_A)
\]
Substituting the values:
\[
0.983 = 0.143 \cdot X_A + 0.84 \cdot (1 - X_A)
\]
### Step 6: Solve for \( X_A \)
Rearranging the equation:
\[
0.983 = 0.143X_A + 0.84 - 0.84X_A
\]
\[
0.983 - 0.84 = (0.143 - 0.84)X_A
\]
\[
0.143X_A - 0.84X_A = 0.143X_A - 0.84X_A = -0.697X_A
\]
\[
0.143 - 0.84 = -0.697X_A
\]
Solving for \( X_A \):
\[
X_A = \frac{0.143 - 0.84}{-0.697}
\]
### Step 7: Calculate Moles of Liquid A
Using the mole fraction \( X_A \):
\[
n_A = X_A \cdot n_T
\]
Where \( n_T \) is the total number of moles, calculated from the mass of water:
\[
n_B = \frac{W_B}{M_B} = \frac{1 \, \text{g}}{18 \, \text{g/mol}} \approx 0.0556 \, \text{mol}
\]
Thus, \( n_T = n_A + n_B \).
### Step 8: Calculate Mass of Liquid A
Using the relationship:
\[
W_A = n_A \cdot M_A
\]
Substituting the values to find \( W_A \).
### Step 9: Final Calculation
Finally, we can find the mass of liquid A per gram of water:
\[
\text{Mass of liquid A per gram of water} = \frac{W_A}{W_B}
\]
