Calculate the mass of urea (`NH_2CONH_2`) required in making 3.5kg of 0.45 molal aqueous solution.

Options

A. 42g
B. 92g
C. 23g
D. 12g

Explanation

To calculate the mass of urea required to make a 3.5 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 per kilogram of solvent. In this case, the solute is urea (NH₂CONH₂) and the solvent is water.

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

### Step 3: Determine the Mass of Water
Since we need to prepare a solution with a total mass of 3.5 kg, we need to determine how much of that mass is water. Assuming the entire mass is the solution, we can denote the mass of water as \( m_{water} \) and the mass of urea as \( m_{urea} \).

Let:
- \( m_{water} = 3.5 \, \text{kg} - m_{urea} \)

### Step 4: Set Up the Equation for Molality
Using the definition of molality:
\[ 0.45 = \frac{n_{urea}}{m_{water} \, (\text{in kg})} \]
Where \( n_{urea} \) is the number of moles of urea.

### Step 5: Calculate the Number of Moles of Urea
Since we know that \( n_{urea} = \frac{m_{urea}}{M_{urea}} \), where \( M_{urea} \) is the molar mass of urea (60 g/mol or 0.060 kg/mol), we can substitute:
\[ 0.45 = \frac{\frac{m_{urea}}{0.060}}{3.5 - m_{urea}} \]

### Step 6: Solve for \( m_{urea} \)
Rearranging the equation gives:
\[ 0.45(3.5 - m_{urea}) = \frac{m_{urea}}{0.060} \]
Multiplying both sides by 0.060:
\[ 0.027(3.5 - m_{urea}) = m_{urea} \]
Expanding this:
\[ 0.0945 - 0.027m_{urea} = m_{urea} \]
Combining like terms:
\[ 0.0945 = m_{urea} + 0.027m_{urea} \]
\[ 0.0945 = 1.027m_{urea} \]
Now, solving for \( m_{urea} \):
\[ m_{urea} = \frac{0.0945}{1.027} \approx 0.092 \, \text{kg} \text{ or } 92 \, \text{g} \]

### Final Answer
The mass of urea required is approximately **92 grams**.

---