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3.What is the molality of ammonia in a solution containing0.85g of NH3 in 100 ML of a liquid of density 0.85gcm3 |
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Answer» Answer: Density = 0.85 GRAMS/cm^3 = 0.85 grams/0.001L = 85grams/0.1L Every 0.1L of that solution WEIGHTS 85grams There are 0.85 grams of NH3 in 100cm^3 = 0.85grams/0.1 solution Mass of water in solution (0.1L) = 85-0.85 =84.15grams =0.08415kg Number of moles of NH3 = Mass/Mw of NH3 = 0.85/17=0.05mol Molality = Number of moles / mass of water in kg = 0.05/0.08415=0.597 Molal Explanation: |
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