Calculate the mass of urea (`NH_2CONH_2`) required in making 4.5kg of 0.45 molal aqueous solution.

Explanation

To calculate the mass of urea (NH₂CONH₂) required to make a 4.5 kg of a 0.45 molal aqueous solution, 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 have a 0.45 molal solution, which means there are 0.45 moles of urea per 1 kg of water.

### Step 2: Calculate the number of moles of urea needed for 4.5 kg of solution
Since the solution is 0.45 molal, we need to find out how many kilograms of water are in 4.5 kg of solution.

Assuming the mass of the solute (urea) is negligible compared to the mass of the solvent (water), we can approximate that:
- The mass of water (solvent) is approximately 4.5 kg (total mass) - mass of urea.

To find the mass of urea, we will first need to calculate how much water is in the solution.

### Step 3: Set up the equation for the mass of the solution
Let the mass of urea be \( m \) grams. Then, the mass of water is approximately \( 4500 - m \) grams.

Using the molality definition:
\[ \text{molality} = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]
We can express this as:
\[ 0.45 = \frac{\frac{m}{60}}{\frac{4500 - m}{1000}} \]
Where \( 60 \) g/mol is the molar mass of urea.

### Step 4: Rearranging the equation
Rearranging gives:
\[ 0.45 \cdot (4500 - m) = \frac{m}{60} \cdot 1000 \]
\[ 0.45 \cdot (4500 - m) = \frac{1000m}{60} \]
\[ 0.45 \cdot 4500 - 0.45m = \frac{1000m}{60} \]

### Step 5: Solving for m
Now, we can solve for \( m \):
\[ 2025 - 0.45m = \frac{1000m}{60} \]
Multiplying through by 60 to eliminate the fraction:
\[ 2025 \cdot 60 - 0.45m \cdot 60 = 1000m \]
\[ 121500 - 27m = 1000m \]
Combine like terms:
\[ 121500 = 1000m + 27m \]
\[ 121500 = 1027m \]
Now, divide both sides by 1027:
\[ m = \frac{121500}{1027} \approx 118.3 \text{ grams} \]

### Final Answer
The mass of urea required to make 4.5 kg of a 0.45 molal aqueous solution is approximately **118.3 grams**.

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