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 find the molecular weight of the solute in the given solution, we can follow these steps:

### Step 1: Calculate the elevation in boiling point (ΔTb)
The boiling point of the solution is given as 100.256°C, and the boiling point of pure water is 100°C.

\[ \Delta T_b = T_{b, \text{solution}} - T_{b, \text{pure water}} = 100.256°C - 100°C = 0.256°C \]

### Step 2: Use the formula for boiling point elevation
The formula for boiling point elevation is given by:

\[ \Delta T_b = i \cdot K_b \cdot m \]

Where:
- \(i\) is the van 't Hoff factor (which is 1 for a non-electrolyte),
- \(K_b\) is the ebullioscopic constant (given as 0.512°C/m),
- \(m\) is the molality of the solution.

### Step 3: Substitute known values into the equation
Since \(i = 1\), we can simplify the equation:

\[ 0.256 = 1 \cdot 0.512 \cdot m \]

### Step 4: Solve for molality (m)
Rearranging the equation to solve for molality:

\[ m = \frac{0.256}{0.512} = 0.5 \, \text{mol/kg} \]

### Step 5: Calculate moles of solute
Molality is defined as moles of solute per kilogram of solvent. We have 400 g of water, which is 0.4 kg. Therefore, we can express the moles of solute as:

\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \Rightarrow \text{moles of solute} = m \cdot \text{mass of solvent in kg} \]

Substituting the values we have:

\[ \text{moles of solute} = 0.5 \, \text{mol/kg} \cdot 0.4 \, \text{kg} = 0.2 \, \text{mol} \]

### Step 6: Calculate the molecular weight of the solute
Molecular weight (M) is defined as the mass of the solute divided by the number of moles of solute:

\[ M = \frac{\text{mass of solute}}{\text{moles of solute}} = \frac{10 \, \text{g}}{0.2 \, \text{mol}} = 50 \, \text{g/mol} \]

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

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