Question

The vapor pressure of water at 25.0 degrees C is 23.8 torr. Determine the mass of glucose C6H12O6 in g is needed to add to 237.8 g of water to change the vapor pressure to 22.8 torr

Answer 1

We know that from Raoult's law (Po-P)/Po = X

Where

Po = vapor pressure of pure water = 23.8 torr

P = vapor pressure of solution = 22.8 torr

X = mole fraction of the solute = n/(n+n')

n = number of moles of solute(glucose) = mass/molar mass

    = m / 180(g/mole)

n' = number of moles of water = mass/molar mass

    = 237.8 g/ 18(g/mole)

    = 13.21 mole

Plug the values we get (23.8-22.8) / 23.8 = (m/180)/[(m/180)+13.21]

                                                   0.042 = m/[m+2377.8]

                                    99.87 + 0.042m = m

                                             0.958 m = 99.87                                   

                                                        m = 104.2 g

So the mass of glucose needed is 104.2g