Vapour pressure of of water at 293K is 17.535 mm Hg. Calculate the vapour pressure of the solution at 293K when 25g of glucose is dissolved in 450g of water ?

Explanation

Raoult's law formula
`(P_(0)-P_(s))/(P_(0))=(n_(s))/(n_(0)-n_(s))`
`n_(s)=("wt")/("GMW")=(25)/(180)=0.1389`
`n_(0)=("wt")/("GMW")=(450)/(18)=25`
`P_(0)=17.535" mm of Hg"`
`(17.535-P_(s))/(17.535)=(0.1384)/(0.1389+25)`
`1-(P_(s))/(17.535)=0.005525`
`(P_(s))/(17.535)=1-0.005525`
`P_(s)=0.9944xx17.535=17.438" mm of Hg."`