Calculate the mass of urea (`NH_2CONH_2`) required in making 3.5kg of 0.25 molal aqueous solution.

Explanation

To calculate the mass of urea (NH₂CONH₂) required to make a 3.5 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 (in this case, urea) per kilogram of solvent (water). A 0.25 molal solution means that there are 0.25 moles of urea in 1 kg of water.

### Step 2: Calculate the number of moles of urea in 3.5 kg of solution
Since the solution is 0.25 molal, we can find out how many moles of urea are present in 3.5 kg of solution.

However, we need to first determine how much of that mass is water. The mass of the solvent (water) in the solution is approximately 3.5 kg, but we need to account for the mass of urea that will be added.

Let’s denote the mass of urea as \( x \) grams. The total mass of the solution will then be \( 3500 + x \) grams.

### Step 3: Set up the equation for molality
Using the definition of molality:
\[ 0.25 = \frac{\text{moles of urea}}{\text{mass of solvent in kg}} \]
The mass of solvent (water) can be expressed as:
\[ \text{mass of solvent} = 3.5 - \frac{x}{1000} \text{ kg} \]
Thus, the number of moles of urea can be expressed as:
\[ \text{moles of urea} = \frac{x}{60} \text{ (since the molar mass of urea is 60 g/mol)} \]

### Step 4: Substitute into the molality equation
Substituting into the molality equation gives:
\[ 0.25 = \frac{\frac{x}{60}}{3.5 - \frac{x}{1000}} \]

### Step 5: Solve for \( x \)
Cross-multiplying gives:
\[ 0.25 \left(3.5 - \frac{x}{1000}\right) = \frac{x}{60} \]
Expanding this:
\[ 0.875 - 0.00025x = \frac{x}{60} \]
Multiplying through by 60 to eliminate the fraction:
\[ 60 \times 0.875 - 0.015x = x \]
\[ 52.5 - 0.015x = x \]
Combining like terms:
\[ 52.5 = x + 0.015x \]
\[ 52.5 = 1.015x \]
Now, solving for \( x \):
\[ x = \frac{52.5}{1.015} \approx 51.7 \text{ grams} \]

### Conclusion
The mass of urea required to make 3.5 kg of a 0.25 molal aqueous solution is approximately **51.7 grams**.

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