The vepur pressure of an aqueous solution of glucose is 750 mm of mercury at `100^(@)C`. Calculate the molatlity and mole fraction of solute.

Explanation

Calculation ofmole fraction of solute
`"Vapour pessure of pure water at "373 K (P_(A)^(@))=760mm Hg`
`"Vapour pessure of aqueous solution at " 373 K (P)=750 mm Hg `
`" According to Raoult's Law," P=P_(A)^(@)x_(A) or x_(A)=P/P_(A)^(@)`
`x_(A)=((750mmHg))/((760mmHg))=00.987`
`"Mpe fracion of solute" (x_(B))=1-x_(A)=1-00.987=0.013`
Calculation of molality of solution.
According to available information,
1 , mole of the solution contains 0.013 mole of solute (glucose) and 0.987 mole of solvent (water)
`"Mass of 0.987 mole of water"=0.987 molxx(18 g mol ^(-1))=17.766 g=17,766xx10^(-3)`
`"Molality of solution (m)"=("No.of moles of glucose")/("Mass of solvent in kg")`
` =((0.013 mol))/((17.766xx10^(-3)kg))=0.73 mol kg^(-1)=0.73 m`