A solution containing 10 g of a non- voltile, higher nonelectrolyte and 400 g of water boils at `100.256^(@)C` at 1 atm. The molecular weight of the solute (in g/mol) is
Given : `(K_(b)" for water "0.512^(@)C//m)`

Answer

To solve the problem, we need to find the molecular weight of the solute using the given boiling point elevation data. We will use the formula for boiling point elevation, which is based on colligative properties.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of solute (non-volatile, non-electrolyte) = 10 g
- Mass of solvent (water) = 400 g
- Boiling point of the solution = 100.256°C
- Boiling point of pure water = 100°C
- Kb (boiling point elevation constant for water) = 0.512°C/m

2. **Calculate the Elevation in Boiling Point (ΔTb):**
\[ \Delta T_b = T_b(\text{solution}) - T_b(\text{pure water}) = 100.256°C - 100°C = 0.256°C \]

3. **Use the Boiling Point Elevation Formula:**
The formula for boiling point elevation is:
\[ \Delta T_b = i \cdot K_b \cdot m \]
Where:
- \(i\) = van 't Hoff factor (for non-electrolytes, \(i = 1\))
- \(K_b\) = boiling point elevation constant
- \(m\) = molality of the solution

4. **Substituting the Known Values:**
\[ 0.256 = 1 \cdot 0.512 \cdot m \]
From this, we can solve for molality \(m\):
\[ m = \frac{0.256}{0.512} = 0.5 \, \text{mol/kg} \]

5. **Calculate the Number of Moles of Solute:**
Molality is defined as moles of solute per kilogram of solvent. Since we have 400 g of water (0.4 kg):
\[ m = \frac{\text{moles of solute}}{0.4 \, \text{kg}} \implies \text{moles of solute} = m \cdot 0.4 = 0.5 \cdot 0.4 = 0.2 \, \text{moles} \]

6. **Calculate the Molecular Weight of the Solute:**
The molecular weight (MM) can be calculated using the formula:
\[ \text{MM} = \frac{\text{mass of solute}}{\text{moles of solute}} = \frac{10 \, \text{g}}{0.2 \, \text{moles}} = 50 \, \text{g/mol} \]

### Final Answer:
The molecular weight of the solute is **50 g/mol**.