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

Explanation

Molality `= (w)/(m xx W)xx1000`
where, w= mass of solute in grams ,
W = mass of solvent in grams
`1.0 = (w)/(m xx W)xx1000`
or `(w)/(mxxW)=(1.0)/(1000)=0.001`
Applying Raoult's law for dilute solution,
`(p_(0)-p_(s))/(p_(0))=(w)/(mxxW)xxM(M=18)`
`(760-p_(s))/(760)=0.001xx18 (p_(0)=760"mm at"100^(@)C)`
or `p_(s)=760-760xx0.001xx18`
`=760-13.68`
`=746-32` mm.