The vapour pressure of an aqueous solution of glucose is `750 mm` of `Hg` at `373 K`. Calculate molality and mole fraction of solute.

Answer

0.741 m,0.013

Explanation

Glucose-B, Water-A
`(P_(A)^(0)-P_(solution))/(P_(A)^(0))=X_(B)`
`(760-750)/(760)=X_(B)`
`X_(B)=0.013`
`" Let motality=x"`
`thereforen_(B)=x,n_(A)=1000/18=55.56`
`0.013=(x)/(x+55.56)`
`x=0.741`
`therefore"Motality of solution =0.741"`