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Question 1 of 3
34.2 grams of Sucrose (RMM 342.0) are dissolved in 500 grams of water. What is the molality of the Sucrose in the solution? Answer to 4 significant figures.
molal
Mass of SOLVENT (in kgs)
- Find the moles of Sucrose present:
Moles = Mass / RMM = 34.2 / 342.0 = 0.1 moles Sucrose - Convert the mass of water from grams to kilograms: 500 g = 0.5 kg
- Find the molality of Sucrose
Molality of Sucrose = 0.1 moles/kg
0.5
Molality of Sucrose = 0.2 = 0.2000 moles/kg to 4 sig figs
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Question 2 of 3
340 milligrams of Ammonia are dissolved in 250 grams of water. What is the molality of Ammonia in the solution? Answer to 4 significant figures.
Mass of SOLVENT (in kgs)
- Find the moles of Ammonia present:
Moles = Mass IN GRAMS / RMM = 0.34 / 17.0 = 0.02 moles Ammonia - Convert the mass of water from grams to kilograms: 250 g = 0.25 kg
- Find the molality of Ammonia
Molality of Ammonia = 0.02 moles/kg
0.25
Molality of Ammonia = 0.08 = 0.08000 moles/kg to 4 sig figs
Note: the first zero after the decimal point is NOT significant.
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Question 3 of 3
500 grams of solution contains water and Potassium Chloride (RMM 74.56) only. If the mass of Potassium Chloride present is 24.85 grams, what is the molality of Potassium Chloride in the solution. Answer to 4 significant figures.
- Find the mass of solvent ie water, involved.
Mass of water = Total mass of solution - mass of KCl
Mass of water = 500 - 24.85 g = 475.15 g = 0.47515 kg - Find the number of moles of KCl present:
Moles of KCl = Mass / RMM = 24.85 / 74.56 = 0.33328 moles - Find the molality of KCl
Molality of KCl = Number of moles of KCl
Mass of Water (in kgs)
Molality of KCl = 0.33328 = 0.70142 = 0.7014 moles/kg to 4 sig figs
0.47515
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