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

Explanation

`p_(0)=` Vapour pressure of water at 373K = 760 mm Hg
Using Raoult's law in the following form,
`(p_(0)-p_(s))/(p_(s))=(wM)/(Wm)`
or`(760-750)/(750)=(w)/(Wm)xx18`
or `(w)/(Wxxm)=(10)/(750xx18)`
Molality `= (w)/(Wxxm)xx1000=(10xx1000)/(750xx18)=0.74m`
`p_(s)=` Mole fraction of solvent `xx p_(0)` ,
Mole fraction of solvent `= (750)/(760)`
So, Mole fraction of solute `= (1-(750)/(760))=0.0132`.