A solution of solute X in benzene boils at `0.126^(@)C` higher than benzene. What is the molality of the solution ?

Options

A. 0.05
B. 2
C. 1
D. 20

Explanation

To find the molality of the solution, we can use the formula related to the elevation of boiling point, which is a colligative property. Here’s a step-by-step solution:

### Step 1: Understand the given data
We know that the boiling point of the solution is elevated by \( \Delta T_b = 0.126^\circ C \) compared to pure benzene.

### Step 2: Identify the ebullioscopic constant
The ebullioscopic constant (\( K_b \)) for benzene is given as \( 2.5 \, \text{°C kg/mol} \).

### Step 3: Use the formula for elevation in boiling point
The formula relating the elevation in boiling point to molality is given by:

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

Where:
- \( \Delta T_b \) = elevation in boiling point
- \( K_b \) = ebullioscopic constant
- \( m \) = molality of the solution

### Step 4: Rearrange the formula to find molality
We can rearrange the formula to solve for molality (\( m \)):

\[ m = \frac{\Delta T_b}{K_b} \]

### Step 5: Substitute the known values into the equation
Substituting the values we have:

\[ m = \frac{0.126 \, \text{°C}}{2.5 \, \text{°C kg/mol}} \]

### Step 6: Calculate the molality
Now, perform the calculation:

\[ m = \frac{0.126}{2.5} = 0.0504 \, \text{mol/kg} \]

### Step 7: Round the answer
We can round this to two decimal places:

\[ m \approx 0.05 \, \text{mol/kg} \]

### Final Answer
The molality of the solution is approximately \( 0.05 \, \text{mol/kg} \).

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