When `0.6g` of urea dissolved in `100g` of water, the water will boil at `(K_(b)` for water `= 0.52kJ. mol^(-1)` and normal boiling point of water `=100^(@)C)`:

Options

A. `373.052 K`
B. `273.52K`
C. `372.48K`
D. `273.052K`

Explanation

To solve the problem of finding the boiling point of water when 0.6 g of urea is dissolved in 100 g of water, we will follow these steps:

### Step 1: Identify the given data
- Mass of urea (solute) = 0.6 g
- Mass of water (solvent) = 100 g
- Normal boiling point of water (T_b₀) = 100 °C = 373 K
- Boiling point elevation constant (K_b) for water = 0.52 kJ/mol = 520 J/mol (since 1 kJ = 1000 J)

### Step 2: Calculate the number of moles of urea
To find the number of moles of urea, we use the formula:
\[ \text{Moles of urea} = \frac{\text{mass of urea}}{\text{molar mass of urea}} \]
The molar mass of urea (NH₂CONH₂) is approximately 60 g/mol.

\[ \text{Moles of urea} = \frac{0.6 \, \text{g}}{60 \, \text{g/mol}} = 0.01 \, \text{mol} \]

### Step 3: Calculate the molality of the solution
Molality (m) is defined as the number of moles of solute per kilogram of solvent. We need to convert the mass of water from grams to kilograms.

\[ \text{Mass of water} = 100 \, \text{g} = 0.1 \, \text{kg} \]

Now, we can calculate the molality:
\[ m = \frac{\text{moles of urea}}{\text{mass of water in kg}} = \frac{0.01 \, \text{mol}}{0.1 \, \text{kg}} = 0.1 \, \text{mol/kg} \]

### Step 4: Calculate the boiling point elevation (ΔT_b)
Using the formula for boiling point elevation:
\[ \Delta T_b = K_b \times m \]
Substituting the values:
\[ \Delta T_b = 520 \, \text{J/mol} \times 0.1 \, \text{mol/kg} = 52 \, \text{J/kg} \]

### Step 5: Convert the boiling point elevation to Kelvin
Since 1 J = 1 K, we can directly convert this to Kelvin:
\[ \Delta T_b = 0.052 \, \text{K} \]

### Step 6: Calculate the new boiling point (T_b)
Now we can find the new boiling point of the solution:
\[ T_b = T_b₀ + \Delta T_b = 373 \, \text{K} + 0.052 \, \text{K} = 373.052 \, \text{K} \]

### Step 7: Convert the boiling point back to Celsius
To convert from Kelvin to Celsius:
\[ T_b = 373.052 \, \text{K} - 273 = 100.052 \, °C \]

### Final Answer
The boiling point of the water when 0.6 g of urea is dissolved in 100 g of water is approximately **100.052 °C**.

---