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