An aqueous solution is 10.% glucose by mass (d = 1.039 g/mL at 20°C). Calculate its freezing point, boiling point at 1 atm, and osmotic pressure.
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Solutions, Molarity and Intermolecular Forces | 7 mins | 0 completed | Learn |
Henry's Law | 20 mins | 0 completed | Learn Summary |
Calculate Molarity | 27 mins | 0 completed | Learn Summary |
Mass Percent | 10 mins | 0 completed | Learn |
Molality | 15 mins | 0 completed | Learn |
Mole Fraction | 14 mins | 0 completed | Learn |
The Colligative Properties | 45 mins | 0 completed | Learn |
Additional Practice |
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Making Solutions |
Freezing Point Depression |
Colloid |
Additional Guides |
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ppm |
The Freezing Point Depression |
Boiling Point Elevation |
An aqueous solution is 10.% glucose by mass (d = 1.039 g/mL at 20°C). Calculate its freezing point, boiling point at 1 atm, and osmotic pressure.
The mass percent of a solution is given by:
In a 10% by mass glucose (C6H12O6) solution, this means there are 10 g of glucose for every 100 g solution. From this, we can calculate the mass of water (the solvent):
We need to find the molality of the solution since we'll use it to calculate the freezing point depression and boiling point elevation. Recall that molality is:
The molar mass of glucose is 6(12.01 g/mol C) + 12(1.01 g/mol H) + 6(16.00 g/mol O) = 180.18 g/mol. Calculating the molality of the glucose solution:
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