A mixture of an organic liquid `A` and water distilled under one atmospheric pressure at `99.2^(@)C`. How many grams of steam will be condensed to obtain `1.0 g` of liquid `A` in the distillate? (Vapour pressure of water `99.2^(@)C` is `739 mm Hg`. Molecular weight of `A=123`)

Explanation

To solve the problem, we need to determine how many grams of steam will be condensed to obtain 1.0 g of liquid A in the distillate. We will use the given information about the vapor pressures and molecular weights to find the solution step by step.

### Step 1: Determine the total vapor pressure
The total vapor pressure of the solution at 99.2°C is 760 mm Hg (which is the atmospheric pressure).

### Step 2: Calculate the vapor pressure of liquid A
We know the vapor pressure of water at 99.2°C is 739 mm Hg. Therefore, we can calculate the vapor pressure of liquid A (P_A) as follows:
\[ P_A = P_{total} - P_W = 760 \, \text{mm Hg} - 739 \, \text{mm Hg} = 21 \, \text{mm Hg} \]

### Step 3: Use Raoult's Law
According to Raoult's Law, the partial pressures of the components in the solution are proportional to their mole fractions. This can be expressed as:
\[ \frac{P_A}{P_W} = \frac{X_A}{X_W} \]
Where:
- \(P_A\) = vapor pressure of liquid A
- \(P_W\) = vapor pressure of water
- \(X_A\) = mole fraction of liquid A
- \(X_W\) = mole fraction of water

### Step 4: Express mole fractions in terms of weights
We can express the mole fractions in terms of the weights of the components:
\[ \frac{P_A}{P_W} = \frac{\frac{W_A}{M_A}}{\frac{W_W}{M_W}} \]
Where:
- \(W_A\) = weight of liquid A
- \(M_A\) = molecular weight of liquid A (123 g/mol)
- \(W_W\) = weight of water
- \(M_W\) = molecular weight of water (18 g/mol)

### Step 5: Substitute known values
We know \(W_A = 1.0 \, \text{g}\), \(M_A = 123 \, \text{g/mol}\), \(P_A = 21 \, \text{mm Hg}\), and \(P_W = 739 \, \text{mm Hg}\). Substituting these values into the equation gives:
\[ \frac{21}{739} = \frac{\frac{1.0}{123}}{\frac{W_W}{18}} \]

### Step 6: Rearranging to find \(W_W\)
Rearranging the equation to solve for \(W_W\):
\[ W_W = \frac{1.0 \times 18 \times 739}{123 \times 21} \]

### Step 7: Calculate \(W_W\)
Now, we can calculate \(W_W\):
\[ W_W = \frac{1.0 \times 18 \times 739}{123 \times 21} \approx 5.19 \, \text{g} \]

### Conclusion
Thus, the weight of steam that will be condensed to obtain 1.0 g of liquid A in the distillate is approximately **5.19 grams**.

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