The vapour pressure lowering caused by the addition of 100 g of sucrose (molecular mass = 3429 mol-1) to 1000 g of water, if the vapour pressure of water at 25oC is 23.8 mm of Hg is

Number of moles of C12H22O11=100342=0.292 mol,Number of moles of H2O=100018=55.5 mol,Vapour pressure of pure water, po=23.8 mm HgAccording to Raoult's law, ∆ppo-nn+N⇒                        ∆p23.8=0.2920.292+55.5                                  ∆p=23.8×0.29255.79                                         =0.125 mm Hg