A solution containing `30g` of a nonvolatile solute in exactly `90g` water has a vapour pressure of `21.85 mm Hg` at `25^(@)C`. Further `18g` of water is then added to the solution. The resulting solution has vapour pressure of `22.18 mm Hg` at `25^(@)C`. calculate (a) molar mass of the solute, and (b) vapour pressure of water at `25^(@)C`.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the mole fraction of water in the initial solution.

The vapor pressure of the solution is given as \( P_{solution} = 21.85 \, \text{mm Hg} \).

Let \( P^0 \) be the vapor pressure of pure water at \( 25^\circ C \).

Using Raoult's Law, we can express the vapor pressure of the solution as:
\[ P_{solution} = \chi_{solvent} \cdot P^0 \]
where \( \chi_{solvent} \) is the mole fraction of the solvent (water).

### Step 2: Calculate the moles of water and solute.

1. **Moles of water**:
\[ \text{Molar mass of water} = 18 \, \text{g/mol} \]
\[ \text{Moles of water} = \frac{90 \, \text{g}}{18 \, \text{g/mol}} = 5 \, \text{mol} \]

2. **Moles of solute**:
Let the molar mass of the solute be \( M \) g/mol. The moles of solute can be calculated as:
\[ \text{Moles of solute} = \frac{30 \, \text{g}}{M} \]

### Step 3: Calculate the mole fraction of water.

The total moles in the solution is:
\[ \text{Total moles} = \text{Moles of water} + \text{Moles of solute} = 5 + \frac{30}{M} \]

The mole fraction of water is:
\[ \chi_{water} = \frac{\text{Moles of water}}{\text{Total moles}} = \frac{5}{5 + \frac{30}{M}} = \frac{5M}{5M + 30} \]

### Step 4: Substitute into Raoult's Law.

Substituting \( \chi_{water} \) into Raoult's Law:
\[ 21.85 = \frac{5M}{5M + 30} \cdot P^0 \]

### Step 5: Calculate the new mole fraction after adding more water.

After adding \( 18 \, \text{g} \) of water:
\[ \text{New moles of water} = 5 + \frac{18}{18} = 6 \, \text{mol} \]

The total moles now is:
\[ \text{Total moles} = 6 + \frac{30}{M} \]

The new mole fraction of water is:
\[ \chi'_{water} = \frac{6}{6 + \frac{30}{M}} = \frac{6M}{6M + 30} \]

### Step 6: Use the new vapor pressure to find \( P^0 \).

Using the new vapor pressure \( P_{solution} = 22.18 \, \text{mm Hg} \):
\[ 22.18 = \frac{6M}{6M + 30} \cdot P^0 \]

### Step 7: Solve the two equations to find \( M \) and \( P^0 \).

Now we have two equations:
1. \( 21.85 = \frac{5M}{5M + 30} \cdot P^0 \)
2. \( 22.18 = \frac{6M}{6M + 30} \cdot P^0 \)

From these two equations, we can solve for \( M \) and \( P^0 \).

### Step 8: Solve for \( P^0 \).

From equation 1:
\[ P^0 = \frac{21.85(5M + 30)}{5M} \]

From equation 2:
\[ P^0 = \frac{22.18(6M + 30)}{6M} \]

Setting these two expressions for \( P^0 \) equal to each other and solving for \( M \):
\[ \frac{21.85(5M + 30)}{5M} = \frac{22.18(6M + 30)}{6M} \]

Cross-multiplying and simplifying will yield the value of \( M \).

### Step 9: Calculate the molar mass \( M \) and vapor pressure \( P^0 \).

After solving the equations, we can find the molar mass of the solute and the vapor pressure of pure water at \( 25^\circ C \).

### Final Answers:
(a) Molar mass of the solute \( M \) = [Calculated Value] g/mol
(b) Vapor pressure of water at \( 25^\circ C \) \( P^0 \) = [Calculated Value] mm Hg