Calculate the mass of urea (`NH_2CONH_2`) required in making 3.5kg of 0.35 molal aqueous solution.

Options

A. 71.9g
B. 74g
C. 35g
D. 26g

Explanation

To calculate the mass of urea (NH₂CONH₂) required to make a 3.5 kg of a 0.35 molal aqueous solution, we can follow these steps:

### Step 1: Understand Molality
Molality (m) is defined as the number of moles of solute per kilogram of solvent. A 0.35 molal solution means there are 0.35 moles of urea in 1 kg of water.

### Step 2: Calculate Moles of Urea in 3.5 kg of Solution
Since the solution is 0.35 molal, we need to determine how many moles of urea are present in the solvent (water) in the 3.5 kg solution.

For a 0.35 molal solution:
- In 1 kg of water, there are 0.35 moles of urea.
- Therefore, in 3.5 kg of solution, the mass of water is approximately 3.5 kg - mass of urea (which we will calculate).

Let's denote the mass of urea as \( m \) grams. The mass of water in the solution will then be \( 3500 - m \) grams.

### Step 3: Set Up the Equation for Molality
Using the definition of molality:
\[ \text{molality} = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}} \]
We can express the moles of urea as:
\[ \text{moles of urea} = \frac{m}{60} \quad \text{(since the molar mass of urea is 60 g/mol)} \]
The mass of solvent (water) in kg is:
\[ \text{mass of solvent} = \frac{3500 - m}{1000} \]
Setting up the equation:
\[ 0.35 = \frac{\frac{m}{60}}{\frac{3500 - m}{1000}} \]

### Step 4: Solve the Equation
Cross-multiplying gives us:
\[ 0.35 \cdot \left(\frac{3500 - m}{1000}\right) = \frac{m}{60} \]
Multiplying both sides by 60, we get:
\[ 21 \cdot (3500 - m) = 1000m \]
Expanding this:
\[ 73500 - 21m = 1000m \]
Combining like terms:
\[ 73500 = 1000m + 21m \]
\[ 73500 = 1021m \]
Now, solving for \( m \):
\[ m = \frac{73500}{1021} \approx 72.0 \text{ grams} \]

### Step 5: Conclusion
The mass of urea required to make 3.5 kg of a 0.35 molal aqueous solution is approximately **72.0 grams**.

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