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Question 1 of 3
What is the mass of water in grams associated with 0.04 moles of Dextrose if the molality of Dextrose in the solution is 0.1000 molal?
grams
Molality of Component
Mass of water = 0.04 g
0.1000
Mass of water = 0.4 kg
Mass of water = 400 g
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Question 2 of 3
The concentration of Potassium Permanganate (RMM 158.0) in water is 0.6300 molal. What mass of water is associated with 3.95 grams of Potassium Permanganate? Answer to 4 significant figures.
Molality of Component
- Find the number of moles of Potassium Permanganate.
Moles of = Mass / RMM = 3.95 / 158.0 = 0.025 moles - Find the required mass of water.
Mass of water = 0.025 g
0.6300
Mass of water = 0.039682 kg
Mass of water = 39.682 g = 39.68 g to 4 sig figs
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Question 3 of 3
You need to prepare a 0.4200 molal solution of Ethanoic Acid in water. If you have 12.01 grams of Ethanoic Acid (RMM 60.05), what is the FINAL mass in grams of the solution prepared? Answer to 2 decimal places.
Molality of Component
- Find the number of moles of Ethanoic Acid.
Moles of Ethanoic Acid = Mass / RMM = 12.01 / 60.05 = 0.2 moles - Find the mass of water
Mass of water = 0.2 g
0.4200
Mass of water = 0.47619 kg = 476.190 g - Total Mass of Solution is:
Mass of Ethanoic Acid + Mass of Water = 12.01 g + 476.190 g
Mass of Ethanoic Acid + Mass of Water = 488.200 g = 488.20 g to 2 dec pl
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