Question

if 4.27 grams of sucrose, C12G22O11, are dissolved in 15.2 grams of water, what will be the poiling point of the resulting solution? Kb for water = 0.512degreesC/m. explain solution with step by step.

Answer 1

We know that ΔT_f = iK_f x m

Where

ΔT _f = elevation in boiling point

        = boiling point of solution – boiling point of pure solvent(water)

        = (T_f - 100) ^oC

K _f = elevation in boiling point constant of water = 0.512 ^oC/m

i = vanthoff’s factor = 1 ( since sucrose is a non-electrolyte)

m = molality of the solution = (mass/molar mass) / Mass of solvent in kg

     = ( 4.27 g / 342(g/mol) / 0.0125 kg

    = 0.998 m

Plug the values we get

ΔT_f = iK_f x m

T_f - 100 = 1x0.512x0.998

T_f = 100+0.51

   = 100.51 ^oC

Therefore the boiling point of the resulting solution is 100.51 ^oC