A solution of `0.450g` of urea (mol.wt 60) in `22.5g` of water showed `0.170^(@)C` of elevation in boiling point, the molal elevation constant of water:

Options

A. `0.51`
B. `0.95`
C. `0.25`
D. `2.25`

Explanation

To find the molal elevation constant (Kb) of water, we can use the formula for boiling point elevation:

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

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

### Step 1: Calculate the moles of solute (urea)

Given:
- Mass of urea = 0.450 g
- Molar mass of urea = 60 g/mol

To find the moles of urea, we use the formula:

\[ \text{Moles of solute} = \frac{\text{mass of solute}}{\text{molar mass of solute}} = \frac{0.450 \, \text{g}}{60 \, \text{g/mol}} = 0.0075 \, \text{mol} \]

### Step 2: Convert the mass of solvent (water) to kilograms

Given:
- Mass of water = 22.5 g

To convert grams to kilograms:

\[ \text{Mass of solvent in kg} = \frac{22.5 \, \text{g}}{1000} = 0.0225 \, \text{kg} \]

### Step 3: Calculate the molality of the solution

Molality (m) is defined as the number of moles of solute per kilogram of solvent:

\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.0075 \, \text{mol}}{0.0225 \, \text{kg}} = 0.3333 \, \text{mol/kg} \]

### Step 4: Substitute values into the boiling point elevation formula

We know:
- \(\Delta T_b = 0.170 \, ^\circ C\)
- \(m = 0.3333 \, \text{mol/kg}\)

Substituting these values into the formula:

\[ 0.170 = K_b \cdot 0.3333 \]

### Step 5: Solve for \(K_b\)

Rearranging the equation to solve for \(K_b\):

\[ K_b = \frac{0.170}{0.3333} \approx 0.51 \, ^\circ C \cdot \text{kg/mol} \]

### Final Answer:

The molal elevation constant of water \(K_b\) is approximately \(0.51 \, ^\circ C \cdot \text{kg/mol}\).

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