Calculate the vapour pressure of an aqueous solution of `1.0` molal glucose solution at `100^(@)C`.

Explanation

Molarlity =`(W_(2))/(Mw_(2)xxW_(1)) xx 1000`
where `W_(2)` = mass of solute in grams,
`W_(1)` = mass of solvent in grams
`1.0 = (W_(2) xx 1000) /(Mw_(2) xx W_(1))`
`(K/W_(1) = K') = 1.0 /1000 = 0.001`
Applying Raoult's law for dilute solution,
`(P^(@) - P_(S)) / P^(@) = W_(2) / (Mw_(2) xx W_(1)) xx Mw_(1)` `[Mw_(1) = 18]`
`(760 - P_(S))/(760) = 0.001 xx 18`
`=760-13.68`
`746.32 mm Hg`