18 g of glucose (C6 H12O6 ) is added to 178.2 g water. The vapour pressure (in torr) for this aqueous solution is

Vapour pressure of water p∘=760 torr Number of moles of glucose = Mass of glucose (in g)  Molecular mass of glucose gmol−1=18g180gmol−1=0.1molNumber of moles of water =178.2g18gmol−1=9.9molTotal number of moles =(0.1+9.9) moles =10molNow, mole fraction of glucose in solution = Change in pressure with respect to initial pressure                                     Δpp∘=Xs=nsns+no=0.110or                         Δp=0.01p∘=0.01×760=7.6 torr ∵ Vapour pressure of solution =(760−7.6)=752.4 torr.