Calculate the mass of urea `(NH_2 CONH_2)` required in making 2.5 kg of 0.25 molal queous solution.

Answer

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

### Step 1: Calculate the molar mass of urea
The molecular formula of urea is NH₂CONH₂. To find the molar mass, we sum the atomic masses of all the atoms in the formula:
- Nitrogen (N): 2 atoms × 14 g/mol = 28 g/mol
- Hydrogen (H): 4 atoms × 1 g/mol = 4 g/mol
- Carbon (C): 1 atom × 12 g/mol = 12 g/mol
- Oxygen (O): 1 atom × 16 g/mol = 16 g/mol

Adding these together:
\[ \text{Molar mass of urea} = 28 + 4 + 12 + 16 = 60 \, \text{g/mol} \]

### Step 2: Understand the definition of molality
Molality (m) is defined as the number of moles of solute per kilogram of solvent. The formula for molality is:
\[ m = \frac{\text{Number of moles of solute}}{\text{Mass of solvent (kg)}} \]

### Step 3: Set up the equation for molality
Let the mass of urea required be \( w \) grams. The number of moles of urea can be calculated using its molar mass:
\[ \text{Number of moles of urea} = \frac{w}{60} \]

Given that the mass of the solvent (water) is 2.5 kg, we can express the molality as:
\[ 0.25 = \frac{\frac{w}{60}}{2.5} \]

### Step 4: Rearrange the equation to solve for \( w \)
We can rearrange the equation to isolate \( w \):
\[ 0.25 = \frac{w}{60 \times 2.5} \]
\[ 0.25 = \frac{w}{150} \]
Multiplying both sides by 150 gives:
\[ w = 0.25 \times 150 \]
\[ w = 37.5 \, \text{grams} \]

### Conclusion
The mass of urea required to make 2.5 kg of a 0.25 molal aqueous solution is **37.5 grams**.

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