How many grams of sucrose must be added to 552 grams of water to give a solution?

1.35x10^-5 grams of sucrose must be added to produce a solution with vapor pressure of 15.5 mmHg.

Solution:

We can use the Raoult's law to get the mole fraction of water Xsolvent present in the solution:

Psolution = Xsolvent Posolvent

(17.5 mmHg - 2.0 mmHg) = Xsolvent (17.5 mmHg)

Xsolvent = 0.886

We use the molar mass of water to compute for its number of moles:

moles of solvent = (552 g H2O) (1 mol H2O / 18.015 g H2O)

= 30.64 mol H2O

From the mole fraction of the water given by

Xsolvent = moles of solvent / sum of moles of solution

= moles of H2O / (moles of sucrose + moles of H2O)

0.886 = (30.64 mol) / (moles of sucrose + 30.64 mol)

we can solve for the number of moles of sucrose:

moles of solute = (30.64 mol / 0.886) - 30.64 mol = 3.94 mol sucrose

We can now calculate for the mass of sucrose using its molar mass:

mass of sucrose = (3.94 mol sucrose) (342.3 g sucrose / 1 mol sucrose)

= 1.35x10^-5 grams