The vapour pressure of a dilute aqueous solution of glucosse `(C_(6)H_(12)O_(6))` is `750 mm Hg` at `273 K`. Calculate (a) molality and (b) mole fraction of the solute.

Answer

`m=0.730`;`chi_(B)=0.013`

Explanation

Given that : Temperature =`273 K`
Boiling point of `H_(2)O=373 K`
Vapour pressure of `H_(2)O=76 cm`
we have,
`(P^(@)-P_(S))/P^(@) =(W_(2)xxMw_(1))/(Mw_(2) xx W_(1))`
`:. "Molality" =W_(2)/(Mw_(2) xx W_(1))xx 1000 =(P^(@)-P_(S))/P^(@) xx1/(Mw_(1)) xx1000`
=`(760-750)/760 xx1/18 xx100`
`=0.730 "mol"//"kg of solvent"`
Also we have
`(P^(@)-P_(S))/P^(@) =n_(2)/(n_(2)+n_(1))`
`:. "Mole fraction" =(P^(@)-P_(S))/P^(@)=(760-750)/760=10/760 =0.013`