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 vapor pressures of water and the organic liquid A, along with their molecular weights, to find the solution.

### Step-by-Step Solution:

1. **Identify Given Data:**
- Vapor pressure of water at 99.2°C, \( P_A = 739 \, \text{mm Hg} \)
- Total atmospheric pressure, \( P_{total} = 760 \, \text{mm Hg} \)
- Molecular weight of liquid A, \( M_A = 123 \, \text{g/mol} \)
- Mass of liquid A in distillate, \( m_A = 1.0 \, \text{g} \)

2. **Calculate Vapor Pressure of Liquid A:**
- Using Dalton's law of partial pressures, we can find the vapor pressure of liquid A, \( P_B \):
\[ P_B = P_{total} - P_A = 760 \, \text{mm Hg} - 739 \, \text{mm Hg} = 21 \, \text{mm Hg} \]

3. **Calculate Mole Fraction of Liquid A:**
- The mole fraction of liquid A in the mixture can be calculated using the formula:
\[ \frac{P_A}{P_B} = \frac{n_A}{n_B} \]
- Rearranging gives:
\[ \frac{739}{21} = \frac{n_A}{n_B} \]

4. **Express Moles in Terms of Mass:**
- The number of moles can be expressed as:
\[ n_A = \frac{m_A}{M_A} = \frac{1.0 \, \text{g}}{123 \, \text{g/mol}} \approx 0.00813 \, \text{mol} \]
- Let \( n_B \) be the number of moles of water. We can express \( n_B \) in terms of the mole fraction:
\[ n_B = n_A \cdot \frac{P_B}{P_A} = 0.00813 \cdot \frac{21}{739} \approx 0.00023 \, \text{mol} \]

5. **Calculate Mass of Water (Steam) Condensed:**
- The mass of water (steam) condensed can be calculated using the number of moles:
\[ m_B = n_B \cdot M_B \]
- Where \( M_B \) (molar mass of water) is \( 18 \, \text{g/mol} \):
\[ m_B = 0.00023 \, \text{mol} \cdot 18 \, \text{g/mol} \approx 0.00414 \, \text{g} \]

6. **Calculate Total Mass of Steam Condensed:**
- Since we need to find out how much steam condenses to give 1.0 g of liquid A, we can use the ratio of moles:
\[ \text{Total mass of steam} = \frac{m_B}{m_A} \cdot m_A = \frac{0.00414}{1.0} \cdot 1.0 \approx 0.00414 \, \text{g} \]

7. **Final Calculation:**
- To find the total mass of steam condensed for 1 g of liquid A:
\[ \text{Total mass of steam} = \frac{m_A}{m_B} = \frac{1.0 \, \text{g}}{0.00414 \, \text{g}} \approx 241.5 \, \text{g} \]

### Final Answer:
Approximately **241.5 grams** of steam will be condensed to obtain 1.0 g of liquid A in the distillate.