Calculate the mass of urea `(NH_(2)CONH_(2))` required in 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 calculate 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: Understand the definition of molality
Molality (m) is defined as the number of moles of solute per kilogram of solvent. In this case, we need to find out how many moles of urea are needed for a 0.25 molal solution.

### Step 2: Calculate the number of moles of urea
Given that the molality is 0.25 mol/kg and the mass of the solvent (water) is 2.5 kg, we can calculate the number of moles of urea required:

\[ \text{Number of moles of urea} = \text{molality} \times \text{mass of solvent (in kg)} \]

\[ \text{Number of moles of urea} = 0.25 \, \text{mol/kg} \times 2.5 \, \text{kg} = 0.625 \, \text{mol} \]

### Step 3: Calculate the molar mass of urea
The molar mass of urea (NH₂CONH₂) can be calculated as follows:

- Nitrogen (N): 14 g/mol × 2 = 28 g/mol
- Carbon (C): 12 g/mol × 1 = 12 g/mol
- Oxygen (O): 16 g/mol × 1 = 16 g/mol
- Hydrogen (H): 1 g/mol × 4 = 4 g/mol

Adding these together:

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

### Step 4: Calculate the mass of urea required
Now that we have the number of moles and the molar mass, we can calculate the mass of urea needed:

\[ \text{Mass of urea} = \text{Number of moles} \times \text{Molar mass} \]

\[ \text{Mass of urea} = 0.625 \, \text{mol} \times 60 \, \text{g/mol} = 37.5 \, \text{g} \]

### Step 5: Round the answer
Since we typically round to two significant figures based on the given data, we can round 37.5 g to 37 g.

### Final Answer
The mass of urea required to make 2.5 kg of a 0.25 molal aqueous solution is **37 g**.

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