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

Options

A. 0.64
B. 0.741
C. 0.68
D. 0.94

Explanation

`(P^(@)-P_(s))/(P^(s))=(w_(1)xxm_(2))/(m_(1)xxw_(2))` (Assuming the solution to be dilute and remembering that at 373K the equilibrium vapour pressure of water is 760 mm of Hg).
`implies(P^(@)-P_(s))/(P_(s))=((w_(1)xx1000))/((m_(1)xxw_(2)))xx(m_(2))/1000xx(m_(2))/1000=Mxx(m_(2))/(1000implies(760-750)/750)=Mxx18/1000`
`M=(10xx1000)/(750xx18)=0.741` mol/kg of solvent.