The boiling point of a glucose solution containing 12 g of glucose in 100 g of water is `100.34^@C`. Boiling point of water is `100^@C`. The molal elevation constant of water is

Options

A. `0.51^@C//"Molal"`
B. `51^@C//"Molal"`
C. `5.1^@C//"Molal"`
D. None of these

Answer

To find the molal elevation constant (Kb) of water based on the given data, we can follow these steps:

### Step 1: Determine the elevation in boiling point (ΔTb)
The formula for the elevation in boiling point is:
\[ \Delta T_b = T_b(\text{solution}) - T_b(\text{solvent}) \]
Given:
- \(T_b(\text{solution}) = 100.34^\circ C\)
- \(T_b(\text{solvent}) = 100^\circ C\)

Calculating ΔTb:
\[ \Delta T_b = 100.34 - 100 = 0.34^\circ C \]

### Step 2: Calculate the molality (m) of the solution
Molality is defined as:
\[ m = \frac{\text{mass of solute (kg)}}{\text{molar mass of solute (g/mol)} \times \text{mass of solvent (kg)}} \]
Given:
- Mass of glucose (solute) = 12 g
- Molar mass of glucose (C6H12O6) = \(6 \times 12 + 12 \times 1 + 6 \times 16 = 180 \, g/mol\)
- Mass of water (solvent) = 100 g = 0.1 kg

Calculating molality:
\[ m = \frac{12 \, g}{180 \, g/mol \times 0.1 \, kg} = \frac{12}{180 \times 0.1} = \frac{12}{18} = 0.67 \, mol/kg \]

### Step 3: Use the boiling point elevation formula to find Kb
The formula relating boiling point elevation to molality and Kb is:
\[ \Delta T_b = m \times K_b \]
Rearranging for Kb gives:
\[ K_b = \frac{\Delta T_b}{m} \]

Substituting the values:
\[ K_b = \frac{0.34^\circ C}{0.67 \, mol/kg} \approx 0.507 \,^\circ C \, mol^{-1} \, kg \]

### Step 4: Round to appropriate significant figures
Rounding the value gives:
\[ K_b \approx 0.51^\circ C \, mol^{-1} \, kg \]

### Final Answer
The molal elevation constant of water is approximately \(0.51^\circ C \, mol^{-1} \, kg\).

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