The solution of 2.5 g of a non-volatile substance in 100 g of benzene boiled at a temperature `0.42^(@)C` higher than the b.p. of pure benzene. Calculate mol. Wt. of the substance. (`K_(b)` of benzene is 2.67 K kg `"mole"^(-1)`)

Explanation

To solve the problem of calculating the molecular weight of a non-volatile substance dissolved in benzene, we will follow these steps:

### Step 1: Understand the Given Data
- Mass of the non-volatile substance (solute), \( W = 2.5 \, \text{g} \)
- Mass of benzene (solvent), \( W' = 100 \, \text{g} \)
- Elevation in boiling point, \( \Delta T_b = 0.42 \, \text{°C} \)
- Boiling point elevation constant for benzene, \( K_b = 2.67 \, \text{K kg mole}^{-1} \)

### Step 2: Convert Mass of Benzene to Kilograms
Since molality is expressed in kg of solvent, we need to convert the mass of benzene from grams to kilograms:
\[ W' = 100 \, \text{g} = 0.1 \, \text{kg} \]

### Step 3: Use the Boiling Point Elevation Formula
The formula for boiling point elevation is given by:
\[ \Delta T_b = K_b \times m \]
where \( m \) is the molality of the solution. We can express molality as:
\[ m = \frac{\text{moles of solute}}{\text{kg of solvent}} = \frac{\frac{W}{M}}{W'} \]
where \( M \) is the molecular weight of the solute.

### Step 4: Substitute the Molality into the Elevation Formula
Substituting the expression for molality into the boiling point elevation formula:
\[ \Delta T_b = K_b \times \frac{\frac{W}{M}}{W'} \]
Rearranging gives:
\[ \Delta T_b = \frac{K_b \times W}{M \times W'} \]

### Step 5: Rearranging to Find Molecular Weight
Now, we can rearrange this equation to solve for \( M \):
\[ M = \frac{K_b \times W}{\Delta T_b \times W'} \]

### Step 6: Substitute the Known Values
Now we can substitute the known values into the equation:
\[ M = \frac{2.67 \, \text{K kg mole}^{-1} \times 2.5 \, \text{g}}{0.42 \, \text{°C} \times 0.1 \, \text{kg}} \]

### Step 7: Calculate
Calculating the numerator:
\[ 2.67 \times 2.5 = 6.675 \]
Calculating the denominator:
\[ 0.42 \times 0.1 = 0.042 \]
Now, substituting these values:
\[ M = \frac{6.675}{0.042} \approx 158.93 \, \text{g/mol} \]

### Final Answer
The molecular weight of the non-volatile substance is approximately \( 158.93 \, \text{g/mol} \).

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