What weight of solute (mol. Wt. 60) is required to dissolve in 180 g of water to reduce the vapour pressure to `4//5^(th)` of pure water ?

Explanation

To solve the problem, we need to determine the weight of a solute (with a molecular weight of 60 g/mol) that must be dissolved in 180 g of water to reduce the vapor pressure of the solution to \( \frac{4}{5} \) of the vapor pressure of pure water.

### Step-by-Step Solution:

1. **Understand the Given Information:**
- Molecular weight of the solute = 60 g/mol
- Mass of water (solvent) = 180 g
- Final vapor pressure of the solution = \( \frac{4}{5} \) of the vapor pressure of pure water.

2. **Apply Raoult's Law:**
- According to Raoult's Law, the relative lowering of vapor pressure is equal to the mole fraction of the solute.
- If \( P_0 \) is the vapor pressure of pure water, then the vapor pressure of the solution is \( P = \frac{4}{5} P_0 \).
- The lowering of vapor pressure, \( \Delta P = P_0 - P = P_0 - \frac{4}{5} P_0 = \frac{1}{5} P_0 \).

3. **Set Up the Equation:**
- The relative lowering of vapor pressure can be expressed as:
\[ \frac{\Delta P}{P_0} = \frac{1}{5} \]
- This is equal to the mole fraction of the solute (\( X_{solute} \)):
\[ X_{solute} = \frac{n_{solute}}{n_{solute} + n_{solvent}} = \frac{1}{5} \]

4. **Calculate Moles of Solvent:**
- The number of moles of water (solvent) can be calculated using its mass and molecular weight:
\[ n_{solvent} = \frac{mass_{water}}{molecular\ weight_{water}} = \frac{180\ g}{18\ g/mol} = 10\ mol \]

5. **Express Mole Fraction in Terms of Moles of Solute:**
- Let \( x \) be the mass of the solute. The number of moles of solute is:
\[ n_{solute} = \frac{x}{60} \]
- The mole fraction equation becomes:
\[ \frac{\frac{x}{60}}{\frac{x}{60} + 10} = \frac{1}{5} \]

6. **Cross Multiply and Solve for \( x \):**
- Cross multiplying gives:
\[ 5 \cdot \frac{x}{60} = \frac{x}{60} + 10 \]
- Simplifying this results in:
\[ \frac{5x}{60} = \frac{x}{60} + 10 \]
- Multiply through by 60 to eliminate the denominator:
\[ 5x = x + 600 \]
- Rearranging gives:
\[ 5x - x = 600 \implies 4x = 600 \implies x = 150 \]

7. **Conclusion:**
- The weight of the solute required is \( \boxed{150\ g} \).