Calculate the mass of urea (`NH_2CONH_2`) required in making 2.5kg of 0.45 molal aqueous solution.

Explanation

To calculate the mass of urea (NH₂CONH₂) required to make a 2.5 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 in 1 kg of water.

### Step 2: Calculate the number of moles of urea needed for 2.5 kg of solution
Since the solution is 0.45 molal, we need to find out how many moles of urea are required for 2.5 kg of solution.

First, we need to determine the mass of the solvent (water) in the solution. Since the total mass of the solution is 2.5 kg and we know that the mass of the solvent is 1 kg for 0.45 moles of urea, we can set up a proportion.

Let \( x \) be the mass of urea in grams. The mass of the solvent (water) will then be:
\[ \text{Mass of solvent} = 2.5 \text{ kg} - \text{mass of urea} \]

### Step 3: Set up the equation using molality
Using the definition of molality:
\[ 0.45 = \frac{\text{moles of urea}}{\text{mass of solvent in kg}} \]

For 2.5 kg of solution:
\[ \text{mass of solvent} = 2.5 - \frac{x}{1000} \text{ kg} \]

### Step 4: Calculate moles of urea
We can express the moles of urea as:
\[ \text{moles of urea} = \frac{x}{60} \]
where 60 g/mol is the molar mass of urea.

### Step 5: Substitute and solve for x
Substituting into the molality equation:
\[ 0.45 = \frac{\frac{x}{60}}{2.5 - \frac{x}{1000}} \]

Now, cross-multiply to solve for \( x \):
\[ 0.45 \left(2.5 - \frac{x}{1000}\right) = \frac{x}{60} \]

Multiply through by 60, to eliminate the fraction:
\[ 27 \left(2.5 - \frac{x}{1000}\right) = x \]

Distributing:
\[ 67.5 - \frac{27x}{1000} = x \]

### Step 6: Rearrange and solve for x
Rearranging gives:
\[ 67.5 = x + \frac{27x}{1000} \]

Combining like terms:
\[ 67.5 = x \left(1 + \frac{27}{1000}\right) \]
\[ 67.5 = x \left(\frac{1000 + 27}{1000}\right) \]
\[ 67.5 = x \left(\frac{1027}{1000}\right) \]

Now, solving for \( x \):
\[ x = \frac{67.5 \times 1000}{1027} \]
\[ x \approx 65.72 \text{ grams} \]

### Final Answer
The mass of urea required to make 2.5 kg of a 0.45 molal aqueous solution is approximately **65.72 grams**.

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