Calculate the mass of urea (`NH_2CONH_2`) required in making 4.5kg of 0.35 molal aqueous solution.

Explanation

To calculate the mass of urea (NH₂CONH₂) required to make a 4.5 kg of a 0.35 molal aqueous solution, 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). A 0.35 molal solution means there are 0.35 moles of urea in 1 kg of water.

### Step 2: Calculate the number of moles of urea required for 4.5 kg of solution
Since the solution is 0.35 molal, we first need to find out how many kilograms of water are present in 4.5 kg of solution.

Assuming that the mass of the solute (urea) is negligible compared to the mass of the solvent, we can approximate that the mass of the solvent (water) is about 4.5 kg.

However, to be more accurate, we will consider the mass of the urea that will be added to the solution.

### Step 3: Calculate the mass of water in the solution
Let \( x \) be the mass of urea in grams. The mass of water in the solution will then be:
\[ \text{Mass of water} = 4500 \, \text{g} - x \]

### Step 4: Set up the equation for molality
Using the definition of molality:
\[ 0.35 = \frac{\text{moles of urea}}{\text{mass of water in kg}} \]
The moles of urea can be calculated as:
\[ \text{moles of urea} = \frac{x}{60} \]
where 60 g/mol is the molar mass of urea.

The mass of water in kg is:
\[ \text{mass of water in kg} = \frac{4500 - x}{1000} \]

### Step 5: Substitute and solve for \( x \)
Now substituting into the molality equation:
\[ 0.35 = \frac{\frac{x}{60}}{\frac{4500 - x}{1000}} \]
Cross-multiplying gives:
\[ 0.35 \times (4500 - x) = \frac{x}{60} \times 1000 \]
\[ 0.35 \times 4500 - 0.35x = \frac{1000x}{60} \]
\[ 1575 - 0.35x = \frac{1000x}{60} \]
Now, multiplying through by 60 to eliminate the fraction:
\[ 60 \times 1575 - 21x = 1000x \]
\[ 94500 - 21x = 1000x \]
Combining like terms:
\[ 94500 = 1021x \]
Solving for \( x \):
\[ x = \frac{94500}{1021} \approx 92.5 \, \text{g} \]

### Final Answer
The mass of urea required to make 4.5 kg of a 0.35 molal aqueous solution is approximately **92.5 grams**.

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