What is the mass of urea required for making 2.5 kg of 0.25 molal aqueous solution?

Options

A. 37 g
B. 25 g
C. 125 g
D. 27.5 g

Explanation

To find the mass of urea required for making a 2.5 kg of 0.25 molal aqueous solution, we can follow these steps:

### Step 1: Understand the definition of molality
Molality (m) is defined as the number of moles of solute per kilogram of solvent. A 0.25 molal solution means there are 0.25 moles of urea per 1 kg of solvent.

### Step 2: Calculate the number of moles of urea needed for 2.5 kg of solvent
Since we have 2.5 kg of solvent, we can calculate the moles of urea required:
\[ \text{Moles of urea} = \text{Molality} \times \text{Mass of solvent (kg)} = 0.25 \, \text{mol/kg} \times 2.5 \, \text{kg} = 0.625 \, \text{moles} \]

### Step 3: Determine the molecular weight of urea
The molecular formula of urea is \( \text{CH}_4\text{N}_2\text{O} \). To find its molecular weight:
- Carbon (C): 1 atom × 12 g/mol = 12 g/mol
- Hydrogen (H): 4 atoms × 1 g/mol = 4 g/mol
- Nitrogen (N): 2 atoms × 14 g/mol = 28 g/mol
- Oxygen (O): 1 atom × 16 g/mol = 16 g/mol

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

### Step 4: Calculate the mass of urea required
Now, we can find the mass of urea needed using the number of moles and the molecular weight:
\[ \text{Mass of urea} = \text{Moles of urea} \times \text{Molecular weight} = 0.625 \, \text{moles} \times 60 \, \text{g/mol} = 37.5 \, \text{g} \]

### Step 5: Round the mass to an appropriate value
Since we typically round to the nearest whole number in practical applications, we can say:
\[ \text{Mass of urea required} \approx 37 \, \text{g} \]

### Final Answer
The mass of urea required for making a 2.5 kg of 0.25 molal aqueous solution is approximately **37 grams**.

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