Calculate elevation in boiling point for 2 molal aqueous solution of glucose (Given: `K_(b(H_2O) =0.5K kg mol^(-1)))`

Options

A. `1^@C`
B. `4^@C`
C. `3^@C`
D. `2^@C`

Answer

To calculate the elevation in boiling point for a 2 molal aqueous solution of glucose, we can follow these steps:

### Step 1: Understand the formula for elevation in boiling point
The elevation in boiling point (\( \Delta T_b \)) can be calculated using the formula:
\[ \Delta T_b = i \cdot K_b \cdot m \]
where:
- \( \Delta T_b \) = elevation in boiling point
- \( i \) = van 't Hoff factor (number of particles the solute breaks into)
- \( K_b \) = ebullioscopic constant of the solvent (water in this case)
- \( m \) = molality of the solution

### Step 2: Identify the values
- For glucose, which is a non-electrolyte, the van 't Hoff factor \( i = 1 \) (it does not dissociate into ions).
- The given molality \( m = 2 \, \text{molal} \).
- The ebullioscopic constant of water \( K_b = 0.5 \, \text{K kg mol}^{-1} \).

### Step 3: Substitute the values into the formula
Now, we can substitute the values into the formula:
\[ \Delta T_b = i \cdot K_b \cdot m \]
\[ \Delta T_b = 1 \cdot 0.5 \, \text{K kg mol}^{-1} \cdot 2 \, \text{molal} \]

### Step 4: Calculate the elevation in boiling point
\[ \Delta T_b = 1 \cdot 0.5 \cdot 2 = 1 \, \text{°C} \]

### Final Answer
The elevation in boiling point for the 2 molal aqueous solution of glucose is \( 1 \, \text{°C} \).

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