When 10.6 g of a non volatile substance is dissolved in 750 g of ether, its boiling point is raised `0.266^(@)C`. The molecular weight of the substance is
(Given : Molal boiling point constant for ether is `2.0^(@)Ckg//mol`)
Report your answer by rounding it up to nearest whole number.

Answer

To find the molecular weight of the non-volatile substance, we can use the formula for boiling point elevation:

\[ M_2 = \frac{K_b \cdot W_2}{\Delta T_b \cdot W_1} \]

Where:
- \( M_2 \) = molecular weight of the solute (substance)
- \( K_b \) = molal boiling point constant of the solvent (ether)
- \( W_2 \) = mass of the solute (non-volatile substance)
- \( \Delta T_b \) = elevation in boiling point
- \( W_1 \) = mass of the solvent (ether)

### Step 1: Identify the given values
- \( K_b = 2.0 \, °C \, kg/mol \)
- \( W_2 = 10.6 \, g = 0.0106 \, kg \) (convert grams to kilograms)
- \( \Delta T_b = 0.266 \, °C \)
- \( W_1 = 750 \, g = 0.750 \, kg \) (convert grams to kilograms)

### Step 2: Substitute the values into the formula
\[ M_2 = \frac{2.0 \, °C \, kg/mol \cdot 0.0106 \, kg}{0.266 \, °C \cdot 0.750 \, kg} \]

### Step 3: Calculate the numerator
\[ \text{Numerator} = 2.0 \cdot 0.0106 = 0.0212 \, °C \, kg \]

### Step 4: Calculate the denominator
\[ \text{Denominator} = 0.266 \cdot 0.750 = 0.1995 \, °C \, kg \]

### Step 5: Divide the numerator by the denominator
\[ M_2 = \frac{0.0212}{0.1995} \approx 0.106 \, kg/mol \]

### Step 6: Convert to grams per mole
\[ M_2 = 0.106 \, kg/mol \times 1000 \, g/kg = 106 \, g/mol \]

### Step 7: Round to the nearest whole number
The molecular weight of the substance is approximately **106 g/mol**.

### Final Answer
**106**

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