A student needs to prepare an aqueous solution of sucrose at a temperature of 20 oC with a vapor pressure of 15.0 mm Hg. How many grams of sucrose (mm = 342 g/mol) does she need if she uses 375 g H2O? (The vapor pressure of water at 20 oC is 17.5 mm Hg.)
To calculate the grams of sucrose needed to prepare the aqueous solution, we can use Raoult's law which states:
P(solution) = X(solvent) * P(solvent)
Where:
P(solution) is the vapor pressure of the solution
X(solvent) is the mole fraction of the solvent
P(solvent) is the vapor pressure of the pure solvent
First, let's calculate the mole fraction of water in the solution.
The number of moles of water (H2O) can be calculated using the given mass and molar mass of water:
moles of H2O = mass of H2O / molar mass of H2O
moles of H2O = 375 g / 18.0 g/mol = 20.833 mol
The mole fraction of water is the moles of water divided by the total moles in the solution:
X(H2O) = moles of H2O / (moles of H2O + moles of sucrose)
Since we're only working with water and sucrose:
X(H2O) = moles of H2O / (moles of H2O)
Next, we can use Raoult's law to find the vapor pressure of the solution:
P(solution) = X(H2O) * P(H2O)
P(solution) = X(H2O) * 17.5 mm Hg
We're given that the vapor pressure of the solution is 15.0 mm Hg, so we can set up the equation:
15.0 mm Hg = X(H2O) * 17.5 mm Hg
Now we can solve for X(H2O):
X(H2O) = 15.0 mm Hg / 17.5 mm Hg
X(H2O) = 0.857
The mole fraction of sucrose can be calculated as:
X(sucrose) = 1 - X(H2O)
X(sucrose) = 1 - 0.857
X(sucrose) = 0.143
Finally, we can calculate the moles of sucrose needed to achieve this mole fraction:
moles of sucrose = X(sucrose) * (moles of H2O + moles of sucrose)
moles of sucrose = 0.143 * (20.833 mol + moles of sucrose)
moles of sucrose = 2.980 mol + 0.143 moles of sucrose
0.857 moles of sucrose = 2.980 mol
moles of sucrose = 2.980 mol / 0.857
moles of sucrose = 3.474 mol
Finally, we can calculate the mass of sucrose needed using the molar mass:
mass of sucrose = moles of sucrose * molar mass of sucrose
mass of sucrose = 3.474 mol * 342 g/mol
mass of sucrose = 1188.108 g
Therefore, the student needs 1188.108 grams of sucrose to prepare the aqueous solution.