Question

How would you prepare a 1.50 molal solution of sucrose in water? At what temperature would the solution begin to boil?

thank you

Answer 1

Since molality is = Number of moles of solute / Mass of solvent in kg

Calculate the weight of sucrose = 1.5 moles in gms

-------------------------------------------- = 342.297 + 171.149 = 513.445 gms.

[ note that one mole of an entitty (in this case sucrose ) is equal to its molecular mass ]

therefore, for preparing 1.5 molal solution of sucrose , C12 H22 O11 ( molecular mass = 342.297 ) we should weigh 1.5 moles ,or 513.45 gms. of sucrose & add it to 1 kg ( or = 1000 gms. ) of water to get a clear solution.

Calculate the elevation in Boiling Point of water using the following relation -

T_b  - T_{b ⁰} = K_b . m   

, where m = molality of solution given as 1.5 , K_b   = Molal Elevation constant of solvent which is 0.52 K kg mol^-1   for water in this case , T_b^o = boiling point of pure solvent ie. water in K = 373.15 K

T_b = boiling point of solution ( ie sucrose solution in this case )

So , T_b  - 373.15 = 0.52 x 1.5 = 0.78

-----------------T_b = 373.93 K

Hence the 1.5m sucrose solution would begin to boil at 373.93K

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