Question

What will be the partial pressure of aqueous 30 % by weight glucose (FW = 180 g/mol) solution at 100oC if it is known that the pressure of water vapors at the same temperature is 760 Torr? Select one: a. 729 torr b. 532 torr c. 792 torr d. 667 torr

Answer 1

According to Raoults law of relative lowering of vapour pressure,

(P0 - P)/P0 = n1/(n1+n2)

P0 = vapour pressure of pure solvent

P = vapour pressure of Solution

n1 = number of moles of solute

n2 = number of moles of solvent

mass of glucose = (30/100) *180 = 54 g

number of moles of glucose = 54/180 = 0.3 moles

mass of water = (100-30)/100 * 18 = 12.6 g

number of moles of water = 12.6/18 = 0.7

(760 - P)/760 = 0.3/(0.3+0.7)

P = partial presssure of solution = 532 torrs