The boiling point of a solution of `0.1050g` of a substance in `15.84g` of ether was found to be `0,1^(@)C`higher than that of pure ether.What is the molecular weight of the substance [Molecular elevation constant of ether is 2.16]`

Options

A. `144.50`
B. `143.18`
C. `140.28`
D. `146.66`

Explanation

To find the molecular weight of the substance, we can use the formula for boiling point elevation, which is given by:

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

Where:
- \(\Delta T_b\) = boiling point elevation (in °C)
- \(K_b\) = ebullioscopic constant (molecular elevation constant)
- \(m\) = molality of the solution (in mol/kg)

### Step 1: Identify the known values
- Mass of the solute (substance) = \(0.1050 \, g\)
- Mass of the solvent (ether) = \(15.84 \, g\)
- Boiling point elevation (\(\Delta T_b\)) = \(0.1 \, °C\)
- Molecular elevation constant of ether (\(K_b\)) = \(2.16 \, °C \cdot kg/mol\)

### Step 2: Convert the mass of the solvent to kg
\[ \text{Mass of ether in kg} = \frac{15.84 \, g}{1000} = 0.01584 \, kg \]

### Step 3: Calculate the molality of the solution
Using the boiling point elevation formula, we can rearrange it to find molality \(m\):
\[ m = \frac{\Delta T_b}{K_b} \]
Substituting the known values:
\[ m = \frac{0.1 \, °C}{2.16 \, °C \cdot kg/mol} \approx 0.0463 \, mol/kg \]

### Step 4: Calculate the number of moles of solute
Using the definition of molality:
\[ m = \frac{n}{\text{mass of solvent in kg}} \]
Where \(n\) is the number of moles of solute. Rearranging gives:
\[ n = m \cdot \text{mass of solvent in kg} \]
Substituting the values:
\[ n = 0.0463 \, mol/kg \cdot 0.01584 \, kg \approx 0.000733 \, mol \]

### Step 5: Calculate the molecular weight of the substance
The molecular weight (M) can be calculated using the formula:
\[ M = \frac{\text{mass of solute}}{n} \]
Substituting the known values:
\[ M = \frac{0.1050 \, g}{0.000733 \, mol} \approx 143.4 \, g/mol \]

### Final Answer
The molecular weight of the substance is approximately **143.4 g/mol**.