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 given problem step by step, we will break it down into two parts as specified in the question.

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

1. **Identify the given values:**
- Mass of solute (W) = 20 g
- Mass of solvent (water) = 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:**
\[ \text{Relative lowering of vapor pressure} = \frac{P₀ - P}{P₀} = \frac{23.76 - 22.41}{23.76} \]
\[ = \frac{1.35}{23.76} \approx 0.0568 \]

3. **Use Raoult's Law to relate the lowering of vapor pressure to mole fraction:**
\[ \text{Relative lowering of vapor pressure} = \text{Mole fraction of solute} = \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{solvent}}} \]

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

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

6. **Substituting into the mole fraction equation:**
\[ 0.0568 = \frac{\frac{20}{M}}{\frac{20}{M} + 5.56} \]

7. **Cross-multiply and solve for M:**
\[ 0.0568 \left(\frac{20}{M} + 5.56\right) = \frac{20}{M} \]
\[ 0.0568 \cdot 5.56 = \frac{20}{M} - 0.0568 \cdot \frac{20}{M} \]
\[ 0.316 = \frac{20(1 - 0.0568)}{M} \]
\[ M = \frac{20 \cdot 0.9432}{0.316} \approx 60 \text{ g/mol} \]

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

1. **Determine the new vapor pressure (P):**
\[ P = \frac{P₀}{2} = \frac{23.76}{2} = 11.88 \text{ mmHg} \]

2. **Calculate the new relative lowering of vapor pressure:**
\[ \text{Relative lowering} = \frac{P₀ - P}{P₀} = \frac{23.76 - 11.88}{23.76} = \frac{11.88}{23.76} \approx 0.5 \]

3. **Set up the equation using the mole fraction:**
\[ 0.5 = \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{solvent}}} \]

4. **Substituting for moles of solute (W/M) and moles of solvent (5.56):**
\[ 0.5 = \frac{\frac{W}{60}}{\frac{W}{60} + 5.56} \]

5. **Cross-multiply and solve for W:**
\[ 0.5 \left(\frac{W}{60} + 5.56\right) = \frac{W}{60} \]
\[ 0.5 \cdot 5.56 = \frac{W}{60} - 0.5 \cdot \frac{W}{60} \]
\[ 2.78 = \frac{W(1 - 0.5)}{60} \]
\[ W = 2.78 \cdot 120 \approx 333.6 \text{ g} \]

### Final Answers:
(a) The molar mass of the solute is approximately **60 g/mol**.
(b) The mass of the solute required to reduce the vapor pressure to half is approximately **333.6 g**.