Final answer:
To find the grams of urea needed to lower the vapor pressure of water by 2.21 mmHg, use Raoult's law to first find the mole fraction of urea, then calculate the moles of urea and convert to grams using urea's molar mass. The number of significant digits must be respected throughout the calculation.
Step-by-step explanation:
To calculate the amount of urea (NH₂₂CO) required to lower the vapor pressure of water by 2.21 mmHg at 30°C, we can use Raoult's law. Raoult's law states that the vapor pressure of a solvent is directly proportional to the mole fraction of the solvent in the solution. The formula for Raoult's law in terms of vapor pressure lowering is ΔP = P₀ - P = X₂ * P₀, where ΔP is the vapor pressure lowering, P₀ is the vapor pressure of pure water, P is the vapor pressure of the solution, and X₂ is the mole fraction of the solute.
First, we need to calculate the mole fraction of urea. The reduction in vapor pressure (ΔP) is 2.21 mmHg. The problem doesn't provide the vapor pressure of water at 30°C, so we cannot directly calculate X₂ without it. If we had the vapor pressure of pure water at 30°C (P₀), we could solve for X₂ using the equation ΔP = X₂ * P₀. After finding X₂, the number of moles of urea (n₂) can be calculated from the mole fraction with the relation X₂ = n₂ / (n₁ + n₂), where n₁ is the number of moles of water.
Once we have n₂, we can use the molar mass of urea, which is 60.06 g/mol, to convert this to grams by multiplying n₂ by the molar mass. This will give us the final weight of urea needed to add to 427g of water to achieve the vapor pressure change. Remember to use the correct number of significant digits throughout the calculation process to ensure accuracy.