Calculate the mass of urea (`NH_2CONH_2`) required in making 1.5kg of 0.35 molal aqueous solution.

Options

A. 30.85g
B. 36g
C. 21g
D. 14g

Explanation

To calculate the mass of urea (NH₂CONH₂) required to make a 1.5 kg of 0.35 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). Here, we have a 0.35 molal solution, which means there are 0.35 moles of urea in 1 kg of water.

### Step 2: Calculate the number of moles of urea needed for 1.5 kg of solution
Since the solution is 0.35 molal, we need to find out how many kilograms of water are in 1.5 kg of solution. The mass of the solvent (water) can be calculated as follows:

- Total mass of solution = mass of solute (urea) + mass of solvent (water)
- Let the mass of urea be \( m \) grams. Then, the mass of water in the solution is \( 1500 - m \) grams.

### Step 3: Set up the equation for molality
Using the definition of molality:
\[ 0.35 = \frac{\text{moles of urea}}{\text{mass of solvent in kg}} \]
Since 1 kg of water corresponds to 0.35 moles of urea, we can express the moles of urea in terms of \( m \):
\[ \text{Moles of urea} = \frac{m}{60} \quad \text{(since the molar mass of urea is 60 g/mol)} \]
And the mass of solvent in kg is:
\[ \text{mass of solvent} = \frac{1500 - m}{1000} \]

### Step 4: Substitute into the molality equation
Substituting these into the molality equation gives:
\[ 0.35 = \frac{m/60}{(1500 - m)/1000} \]

### Step 5: Cross-multiply and solve for \( m \)
Cross-multiplying gives:
\[ 0.35 \times (1500 - m) = \frac{m}{60} \times 1000 \]
\[ 0.35 \times 1500 - 0.35m = \frac{1000m}{60} \]
\[ 525 - 0.35m = \frac{1000m}{60} \]
To eliminate the fraction, multiply through by 60:
\[ 60 \times 525 - 60 \times 0.35m = 1000m \]
\[ 31500 - 21m = 1000m \]
Combine like terms:
\[ 31500 = 1000m + 21m \]
\[ 31500 = 1021m \]
Now, solve for \( m \):
\[ m = \frac{31500}{1021} \approx 30.85 \text{ grams} \]

### Final Answer
The mass of urea required to make 1.5 kg of a 0.35 molal aqueous solution is approximately **30.85 grams**.

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