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.
We know that ΔTf = iKf x m
Where
ΔT f = elevation in boiling point
= boiling point of solution – boiling point of pure solvent(water)
= (Tf - 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
ΔTf = iKf x m
Tf - 100 = 1x0.512x0.998
Tf = 100+0.51
= 100.51 oC
Therefore the boiling point of the resulting solution is 100.51 oC
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