Question

A solution was prepared by dissolving 25.00 g of glucose in 200.0 g water. The molar mass of glucose is 180.16 g/mol. What is the boiling pointof the resulting solution (in °C)? Glucose is a molecular solid that is present as individual molecules in solution.(Kbis 0.51 °C·kg/mol. ) (CH6)

Answer 1

We know that ΔT _b = K_bx m
Where

ΔT _b= elevation in boiling point

        = boiling point of solution - boilinging point of pure solvent

        = T_b -100^oC

K_b = elevation in boiling point constant of water = 0.51 ^oC.kg/mol = 0.51 oC/m

m = molality of the solution

    = ( mass / Molar mass ) / weight of the solvent in Kg

    = (25.00 / 180.16) / 0.200 kg

    = 0.694 m

Plug the values we get

T_b -100 = 0.51x0.694

T_b =100+0.354

    = 100.354 oC