Questions: What is the molality of a solution of 8.75 g urea, CO(NH2)2 dissolved in 600.0 g water?

Solution Steps

The molecular formula of urea is \(\mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2}\). To find the molar mass, we sum the atomic masses of all the atoms in the formula:

• Carbon (C): \(1 \times 12.01 \, \text{g/mol} = 12.01 \, \text{g/mol}\)
• Oxygen (O): \(1 \times 16.00 \, \text{g/mol} = 16.00 \, \text{g/mol}\)
• Nitrogen (N): \(2 \times 14.01 \, \text{g/mol} = 28.02 \, \text{g/mol}\)
• Hydrogen (H): \(4 \times 1.008 \, \text{g/mol} = 4.032 \, \text{g/mol}\)

Adding these together: \[ 12.01 + 16.00 + 28.02 + 4.032 = 60.06 \, \text{g/mol} \]

Using the mass of urea (8.75 g) and its molar mass (60.06 g/mol), we calculate the number of moles: \[ \text{moles of urea} = \frac{8.75 \, \text{g}}{60.06 \, \text{g/mol}} = 0.1456 \, \text{mol} \]

Molality (m) is defined as the number of moles of solute per kilogram of solvent. Here, the mass of water (solvent) is 600.0 g, which is equivalent to 0.6000 kg: \[ \text{molality} = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.1456 \, \text{mol}}{0.6000 \, \text{kg}} = 0.2427 \, \text{mol/kg} \]

Final Answer