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What is the new boiling point of a solution containing 25.0 g C6H10O5 in 250.0 mL of water?
(1 mol C6H10O5 = 162.14 g, Kb for water = 0.512°C/m)
- jacob sLv 71 month agoFavorite Answer
C6H10O5 mass = 25.0 g
1 mol C6H10O5 = 162.14 g,
moles of C6H10O5 = 25/162.14 = 0.1542 mol
Volume of H2O = 250 ml
mass of H2O = 250 ml x 1g = 250g = 0.250Kg
molality = moles solute/mass solvent kg
molality = 0.154mol/.250 kg = 0.6168 m
Kb = 0.512°C/m
delta Tb = Kb x m = 0.512 x 0.6168 = 0.316 C
New boiling point = 100 C + 0.316C = 100.316 C
I hope this helps you!