Twenty grams of a solute are added to `100g` of water at `25^(@)C`. The vapour pressure of pure water is `23.76 mmHg`, the vapour pressure of the solution is `22.41` Torr.
(a) Calculate the molar mass of the solute.
(b) What mass of this solute is required in `100g` of water of reduce the vapour pressure ot one-half the value for pure water?

Explanation

To solve the problem, we will break it down into two parts as per the question.

### Part (a): Calculate the molar mass of the solute.

1. **Given Data:**
- Mass of solute (Ws) = 20 g
- Mass of water (solvent) = 100 g
- Vapor pressure of pure water (P0) = 23.76 mmHg
- Vapor pressure of the solution (P) = 22.41 mmHg

2. **Calculate the relative lowering of vapor pressure:**
\[ \Delta P = P_0 - P = 23.76 \, \text{mmHg} - 22.41 \, \text{mmHg} = 1.35 \, \text{mmHg} \]
\[ \text{Relative lowering of vapor pressure} = \frac{\Delta P}{P_0} = \frac{1.35}{23.76} \]

3. **Calculate the mole fraction of solute:**
\[ \text{Mole fraction of solute} = \frac{n_s}{n_s + n_w} \]
where \( n_s \) is the number of moles of solute and \( n_w \) is the number of moles of water.

4. **Calculate the number of moles of water:**
- Molar mass of water (H2O) = 18 g/mol
\[ n_w = \frac{100 \, \text{g}}{18 \, \text{g/mol}} \approx 5.56 \, \text{mol} \]

5. **Let the molar mass of the solute be M. The number of moles of solute is:**
\[ n_s = \frac{20 \, \text{g}}{M} \]

6. **Substituting into the mole fraction equation:**
\[ \frac{\frac{20}{M}}{\frac{20}{M} + 5.56} = \frac{1.35}{23.76} \]

7. **Cross-multiplying to solve for M:**
\[ 1.35 \left( \frac{20}{M} + 5.56 \right) = 23.76 \cdot \frac{20}{M} \]
\[ 1.35 \cdot 5.56 + \frac{27}{M} = \frac{23.76 \cdot 20}{M} \]
\[ 7.507 + 1.35 \cdot \frac{20}{M} = 23.76 \cdot \frac{20}{M} \]
\[ 7.507 = (23.76 - 1.35) \cdot \frac{20}{M} \]
\[ 7.507 = 22.41 \cdot \frac{20}{M} \]
\[ M = \frac{22.41 \cdot 20}{7.507} \approx 60 \, \text{g/mol} \]

### Part (b): Mass of solute required to reduce the vapor pressure to half of the pure water.

1. **Calculate the new vapor pressure:**
\[ P = \frac{P_0}{2} = \frac{23.76}{2} = 11.88 \, \text{mmHg} \]

2. **Calculate the relative lowering of vapor pressure for this new condition:**
\[ \Delta P = P_0 - P = 23.76 - 11.88 = 11.88 \, \text{mmHg} \]
\[ \text{Relative lowering of vapor pressure} = \frac{11.88}{23.76} \]

3. **Using the same mole fraction formula:**
\[ \frac{n_s}{n_s + n_w} = \frac{11.88}{23.76} \]
where \( n_w \) remains the same (5.56 mol).

4. **Let the new mass of solute be W:**
\[ n_s = \frac{W}{60} \]
\[ \frac{\frac{W}{60}}{\frac{W}{60} + 5.56} = \frac{11.88}{23.76} \]

5. **Cross-multiplying to solve for W:**
\[ 11.88 \left( \frac{W}{60} + 5.56 \right) = 23.76 \cdot \frac{W}{60} \]
\[ 11.88 \cdot 5.56 + \frac{11.88W}{60} = \frac{23.76W}{60} \]
\[ 66.05 + \frac{11.88W}{60} = \frac{23.76W}{60} \]
\[ 66.05 = \left( \frac{23.76 - 11.88}{60} \right) W \]
\[ 66.05 = \frac{11.88}{60} W \]
\[ W = \frac{66.05 \cdot 60}{11.88} \approx 332 \, \text{g} \]

### Final Answers:
(a) Molar mass of the solute = 60 g/mol
(b) Mass of solute required = 332 g