`0.15`g of a subatance dissolved in `15`g of solvent boiled at a temperature higher at `0.216^(@)`than that of the pure solvent. Calculate the molecular weight of the substance. Molal elecation constant for the solvent is `2.16^(@)C`

Options

A. 10.1
B. 100
C. 1.01
D. 1000

Answer

To calculate the molecular weight of the substance, we can use the formula for boiling point elevation:

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

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

### Step 1: Identify the given values
- \(\Delta T_b = 0.216 \, °C\)
- \(K_b = 2.16 \, °C \, kg/mol\)
- Mass of solute = \(0.15 \, g\)
- Mass of solvent = \(15 \, g\)

### Step 2: Convert the mass of the solvent to kilograms
\[ \text{Mass of solvent in kg} = \frac{15 \, g}{1000} = 0.015 \, kg \]

### Step 3: Rearrange the formula to find molality
We can rearrange the boiling point elevation formula to find molality \(m\):
\[ m = \frac{\Delta T_b}{K_b} \]

### Step 4: Substitute the values to calculate molality
\[ m = \frac{0.216 \, °C}{2.16 \, °C \, kg/mol} = 0.1 \, mol/kg \]

### Step 5: Use the definition of molality to find the molecular weight
Molality \(m\) is also defined as:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]
We can express the moles of solute in terms of its mass and molecular weight \(M\):
\[ m = \frac{\frac{\text{mass of solute}}{M}}{\text{mass of solvent in kg}} \]
Substituting the known values:
\[ 0.1 = \frac{\frac{0.15 \, g}{M}}{0.015 \, kg} \]

### Step 6: Solve for the molecular weight \(M\)
Rearranging the equation gives:
\[ 0.1 \cdot 0.015 \cdot M = 0.15 \]
\[ M = \frac{0.15}{0.1 \cdot 0.015} = \frac{0.15}{0.0015} = 100 \, g/mol \]

### Conclusion
The molecular weight of the substance is \(100 \, g/mol\).

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