Calculate the mass of urea (`NH_2CONH_2`) required in making 5.5kg of 0.45 molal aqueous solution.

Options

A. 144.6g
B. 156g
C. 110g
D. 148.6g

Explanation

To solve the problem of calculating the mass of urea required to make a 5.5 kg of a 0.45 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 (urea in this case) per kilogram of solvent (water).

### Step 2: Calculate the Moles of Urea in 1 kg of Water
Given that the solution is 0.45 molal, it means there are 0.45 moles of urea in 1 kg of water.

### Step 3: Calculate the Total Mass of the Solution
The total mass of the solution is the mass of urea plus the mass of water. Since we are considering 1 kg of water, the total mass of the solution when 0.45 moles of urea is added can be calculated.

### Step 4: Calculate the Molar Mass of Urea
The molar mass of urea (NH₂CONH₂) is 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
Thus, the molar mass of urea = 28 + 12 + 16 + 4 = 60 g/mol.

### Step 5: Calculate the Mass of Urea for 0.45 Moles
Using the molar mass, we can find the mass of urea required for 0.45 moles:
Mass of urea = Number of moles × Molar mass = 0.45 moles × 60 g/mol = 27 grams.

### Step 6: Calculate the Mass of Urea in 5.5 kg of Solution
Now, we need to find out how much urea is needed for 5.5 kg of solution. First, we find the total mass of the solution when 27 grams of urea is added to 1 kg of water:
Total mass of solution = Mass of urea + Mass of water = 27 g + 1000 g = 1027 g = 1.027 kg.

### Step 7: Set Up a Proportion to Find the Mass of Urea in 5.5 kg of Solution
We can set up a proportion:
\[ \frac{27 \text{ g of urea}}{1.027 \text{ kg of solution}} = \frac{x \text{ g of urea}}{5.5 \text{ kg of solution}} \]
Cross-multiplying gives:
\[ x = \frac{27 \text{ g} \times 5.5 \text{ kg}}{1.027 \text{ kg}}. \]

### Step 8: Solve for x
Calculating the above expression:
\[ x = \frac{27 \times 5.5}{1.027} \approx 144.6 \text{ grams}. \]

### Final Answer
The mass of urea required to make 5.5 kg of a 0.45 molal aqueous solution is approximately **144.6 grams**.

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