The molality of a urea solution in which 0.0100 g of urea, `[(NH_2)_2 CO]` is added to `0.3000 dm_3` of water at S.T.P. is:

Options

A. `5.55 xx 10^(-4) M`
B. 33.3 M
C. `3.33 xx 10^(-2) M`
D. 0.555 M

Answer

To find the molality of the urea solution, we will follow these steps:

### Step 1: Calculate the molar mass of urea
The molecular formula of urea is \( (NH_2)_2CO \). To calculate its molar mass:
- Nitrogen (N): 14.01 g/mol (2 Nitrogens = 28.02 g/mol)
- Hydrogen (H): 1.01 g/mol (4 Hydrogens = 4.04 g/mol)
- Carbon (C): 12.01 g/mol (1 Carbon = 12.01 g/mol)
- Oxygen (O): 16.00 g/mol (1 Oxygen = 16.00 g/mol)

Adding these together:
\[ \text{Molar mass of urea} = 28.02 + 4.04 + 12.01 + 16.00 = 60.07 \, \text{g/mol} \]

### Step 2: Calculate the number of moles of urea
Using the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
Given mass of urea = 0.0100 g,
\[ \text{Number of moles of urea} = \frac{0.0100 \, \text{g}}{60.07 \, \text{g/mol}} \approx 1.66 \times 10^{-4} \, \text{moles} \]

### Step 3: Calculate the mass of the solvent (water)
We know the volume of water is 0.3000 dm³. The density of water is approximately 1 g/cm³ or 1000 g/dm³. Therefore, the mass of water can be calculated as:
\[ \text{Mass of water} = \text{Volume} \times \text{Density} = 0.3000 \, \text{dm}^3 \times 1000 \, \text{g/dm}^3 = 300.0 \, \text{g} = 0.3000 \, \text{kg} \]

### Step 4: Calculate the molality of the solution
Molality (m) is defined as the number of moles of solute per kilogram of solvent:
\[ \text{Molality} = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}} \]
Substituting the values:
\[ \text{Molality} = \frac{1.66 \times 10^{-4} \, \text{moles}}{0.3000 \, \text{kg}} \approx 5.55 \times 10^{-4} \, \text{mol/kg} \]

### Final Answer
The molality of the urea solution is approximately \( 5.55 \times 10^{-4} \, \text{mol/kg} \).

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