Calculate the mass of urea `(NH_(2)CONH_(2))` required in making `2.5 kg `of `0.25 `molal aqueous solution.

Explanation

To calculate the mass of urea (NH₂CONH₂) required to make a 2.5 kg of 0.25 molal aqueous solution, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Definition of Molality**:
Molality (m) is defined as the number of moles of solute per kilogram of solvent. The formula is:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}} \]

2. **Identify Given Values**:
- Mass of the solution = 2.5 kg = 2500 g
- Molality (m) = 0.25 molal

3. **Calculate the Mass of Solvent**:
Since the solution consists of both solute (urea) and solvent (water), we need to express the mass of the solvent. Let the mass of urea be \( x \) grams. Thus, the mass of the solvent (water) can be expressed as:
\[ \text{mass of solvent} = \text{mass of solution} - \text{mass of solute} = 2500 \, \text{g} - x \]

4. **Convert Mass of Solvent to Kilograms**:
Since molality is defined in terms of kilograms, we convert the mass of the solvent to kilograms:
\[ \text{mass of solvent (kg)} = \frac{2500 - x}{1000} \]

5. **Calculate Moles of Urea**:
The number of moles of urea can be calculated using its molar mass. The molar mass of urea (NH₂CONH₂) is:
- N: 14 g/mol (2 Nitrogens) = 28 g
- H: 1 g/mol (4 Hydrogens) = 4 g
- C: 12 g/mol (1 Carbon) = 12 g
- O: 16 g/mol (1 Oxygen) = 16 g
\[ \text{Molar mass of urea} = 28 + 4 + 12 + 16 = 60 \, \text{g/mol} \]

6. **Set Up the Equation Using Molality**:
Using the molality formula:
\[ 0.25 = \frac{\text{moles of urea}}{\text{mass of solvent (kg)}} \]
The moles of urea can be expressed as:
\[ \text{moles of urea} = \frac{x}{60} \]
Thus, substituting into the molality equation:
\[ 0.25 = \frac{\frac{x}{60}}{\frac{2500 - x}{1000}} \]

7. **Cross Multiply and Solve for x**:
Rearranging gives:
\[ 0.25 \cdot \frac{2500 - x}{1000} = \frac{x}{60} \]
Cross-multiplying:
\[ 0.25 \cdot (2500 - x) \cdot 60 = 1000x \]
Simplifying:
\[ 15(2500 - x) = 1000x \]
\[ 37500 - 15x = 1000x \]
\[ 37500 = 1015x \]
\[ x = \frac{37500}{1015} \approx 36.9 \, \text{g} \]

### Final Answer:
The mass of urea required to make 2.5 kg of a 0.25 molal aqueous solution is approximately **36.9 grams**.