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:

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

Where:
- \(\Delta T_b\) = boiling point elevation (in °C)
- \(i\) = van 't Hoff factor (which is 1 for non-volatile, non-electrolyte solutes)
- \(K_b\) = molal boiling point elevation constant (given as \(2.0 \, °C \cdot kg/mol\))
- \(m\) = molality of the solution

### Step 1: Calculate the molality (m)

Molality is defined as the number of moles of solute per kilogram of solvent. We first need to calculate the number of moles of the solute.

1. Calculate the molality using the formula:

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

Given:
- Mass of solute = \(10.6 \, g\)
- Mass of solvent (ether) = \(750 \, g = 0.750 \, kg\)

### Step 2: Rearranging the boiling point elevation formula

Since we need to find the moles of solute, we can rearrange the boiling point elevation formula:

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

Substituting \(m\):

\[ \Delta T_b = K_b \cdot \left(\frac{\text{moles of solute}}{0.750}\right) \]

### Step 3: Substitute known values into the equation

Now substituting the known values into the equation:

\[ 0.266 = 2.0 \cdot \left(\frac{\text{moles of solute}}{0.750}\right) \]

### Step 4: Solve for moles of solute

Rearranging gives:

\[ \text{moles of solute} = \frac{0.266 \cdot 0.750}{2.0} \]

Calculating this:

\[ \text{moles of solute} = \frac{0.1995}{2.0} = 0.09975 \, \text{moles} \]

### Step 5: Calculate the molecular weight

Molecular weight (M) is calculated using the formula:

\[ M = \frac{\text{mass of solute (g)}}{\text{moles of solute}} \]

Substituting the values we have:

\[ M = \frac{10.6 \, g}{0.09975 \, \text{moles}} \approx 106.4 \, g/mol \]

### Step 6: Round to the nearest whole number

Rounding \(106.4\) to the nearest whole number gives:

\[ M \approx 106 \, g/mol \]

### Final Answer

The molecular weight of the substance is approximately **106 g/mol**.

---