Question

How many grams of sucrose (C12H22O11) must be added to 477 g of water to give a solution with a vapor pressure 0.555 mmHg less than that of pure water at 20 degrees celsius? (The vapor pressure of water at 20 degrees celsius is 17.5 mmHg).

Answer 1

Po = 17.5 mm Hg
P = 17.5 - 0.555 = 16.945 mm Hg

P = P0* mole fraction of water
16.945 = 17.5 * mole fraction of water
mole fraction of water = 0.9683

number of moles of water = mass/ molar mass
                                                       = 477/18
                                                       = 26.5 mol

Let mass of C12H22O11 be m g
Molar mass of C12H22O11 = 342.3 g/mol
number of moles of C12H22O11 = mass / molar mass
                                                                  = m /342.3

use:
mole fraction of water = moles of water / (moles of water + moles of C12H22O11 )
0.9683 = 26.5 / (26.5 + (m /342.3))
(26.5 + (m /342.3)) = 27.37
m /342.3 = 0.87
m = 297.8 g
Answer: 297.8 g