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

Molality `=0.741 "mol"//kg` of solvent, Mole fraction `=0.013`

Explanation

Given that temperature is `373K` and `b.pt.` of `H_(2)O=373K`
We have, `(P^(0)-P_(s))/(P_(s))=(wxxM)/(mxxM)`
Molality `= =(w)/(mxxW)xx1000=(P^(0)-P_(s))/(P_(s))xx(1)/(M)xx1000=(760-750)/(750)xx(1)/(18)xx0.741 "mol"//kg` of solvent
Also we have, `(P^(0)-P_(s))/(P_(s))=(n)/(n+N)`
Mole fraction `(P^(0)-P_(s))/(P^(0))=(760-750)/(760)=(10)/(760)=0.013`