Final answer:
To find the grams of sucrose needed to reduce the vapor pressure of water by 0.953 mmHg at 20°C, we use Raoult's Law to relate the vapor pressure decrease to the mole fraction of sucrose. Calculations will involve the molar mass of sucrose and the number of moles of water to determine the required sucrose in grams.
Step-by-step explanation:
To determine how many grams of sucrose must be added to 483 g of water to give a solution with a vapor pressure 0.953 mmHg less than that of pure water at 20°C, we can use Raoult's Law. Raoult's Law states that the vapor pressure of a solution is directly proportional to the mole fraction of the solvent. We know the vapor pressure of pure water at 20°C is 17.5 mmHg, so the solution's vapor pressure will be 17.5 mmHg - 0.953 mmHg = 16.547 mmHg.
We next determine the mole fraction of sucrose, knowing that the change in vapor pressure is directly related to the mole fraction of solute. The molar mass of sucrose (C12H22O11) is 342.3 g/mol. Let's denote x as the number of moles of sucrose, then:
x = grams of sucrose / molar mass of sucrose
The mole fraction of sucrose is x / (x + moles of H2O), where moles of H2O = 483 g / 18.015 g/mol.
Using the relationship between vapor pressure decrease and mole fraction, we can set up an equation and solve for the grams of sucrose needed. A detailed calculation would follow these steps, ultimately providing the answer in grams with the correct number of significant digits based on the initial data given.