1.

18g of glucose is dissolved in 1000g of water at 300K. At what temperature does this solution boil?(Kb for water is 0.52 K kg/mol.Molar mass of glucose is 180 g/mol, boiling point of water = 273.15 K)

Answer»

Solution :`"Given 18 g of SOLUTE "=OMEGAT""DeltaT_(b)=K_(b)xxm`
`"Molecular mass "C_(6)H_(12)O_(6)=180""DeltaT_(b)=0.52xx0.1`
`omegat"of the solvent 1000g"DeltaT_(b)=0.052K`
`"B.P of Water"=373.15K`
`"w.K.t"DeltaT_(b)=T_(b)-T_(b)^(@)`
`0.052=T_(b)-373.15`
`T_(b)=373.15+0.052`
`"What will boil at "T_(b)=373.202K`


Discussion

No Comment Found

Related InterviewSolutions