Asked by Miley
A solution is prepared by dissolving 17.1grams of sucrose, c12h22o11, in 275grams of h20.
what is the molality of that solution?
I know what formula to use, but can someone please show me all the steps to get the answer. Thanks:)
what is the molality of that solution?
I know what formula to use, but can someone please show me all the steps to get the answer. Thanks:)
Answers
Answered by
DrBob222
If you know the formula to use you should have no trouble getting the answers.
molality = mols/kg solvent.
mols sucrose = g/molar mass
kg solvent = 0.275 kg.
molality = mols/kg solvent.
mols sucrose = g/molar mass
kg solvent = 0.275 kg.
Answered by
scooby doooo
17.1 g sucrose * 1 mol sucrose/ molar mass (342.34 g) = ~0.05 mol sucrose
m = mol of solute/ kg of solvent (water)
m = 0.05 mol sucrose/ 0.275 kg water = 0.18 m
Answer: 0.18 molality
-- Hope this helps future chem students :)
m = mol of solute/ kg of solvent (water)
m = 0.05 mol sucrose/ 0.275 kg water = 0.18 m
Answer: 0.18 molality
-- Hope this helps future chem students :)
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