In: Chemistry
How would you prepare a 1.50 molal solution of sucrose in water? At what temperature would the solution begin to boil?
thank you
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 -
Tb - Tb 0 = Kb . m
, where m = molality of solution given as 1.5 , Kb = Molal Elevation constant of solvent which is 0.52 K kg mol-1 for water in this case , Tbo = boiling point of pure solvent ie. water in K = 373.15 K
Tb = boiling point of solution ( ie sucrose solution in this case )
So , Tb - 373.15 = 0.52 x 1.5 = 0.78
-----------------Tb = 373.93 K
Hence the 1.5m sucrose solution would begin to boil at 373.93K
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