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Chem question
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A commercial ammonia called "0.68 ammonia" has a density of 0.68 g com-3 and contains approximately 34% pf ammonhia by mass(a) calculatew the mass of ammonia present in 1dm3 of the solution of this commercial ammonia(B)calculate the molarity of this solution (c) calculate the volume of distilled... 顯示更多 A commercial ammonia called "0.68 ammonia" has a density of 0.68 g com-3 and contains approximately 34% pf ammonhia by mass (a) calculatew the mass of ammonia present in 1dm3 of the solution of this commercial ammonia (B)calculate the molarity of this solution (c) calculate the volume of distilled water added if the solution has to be diluted to1 mol dm-4
(a) 1dm3 = 1000cm3 1dm3 of the commercial ammonia weighs 0.68gcm-3 x 1000cm3 = 680 g Since the commercial ammonia contains 34% of ammonia by mass, the mass of the ammonia = 680 x 34% = 231.2 g (b) The RAM of the ammonia = 14 + 1 x 4 = 18 No. of mole of ammonia = 231.2 / 18 = 12.8 moles Molarity of ammonia = 12.8 moles / 1 dm3 = 12.8 M (c) Let a be the volume of water added. Hence 12.8 / (a + 1) = 1 12.8 = a +1 a = 11.8 dm3 Therefore, the volume of water added is 11.8 dm3. 2008-07-22 20:58:24 補充： (b) 應該係 14 + 1 x3 = 17 之後 231.2 / 17 = 13.6 13.6 / 1 = 13.6 M至啱 (c) 13.6 / a + 1 = 1 ，答案係 12.6 dm3
其他解答:
Chem question
發問:
A commercial ammonia called "0.68 ammonia" has a density of 0.68 g com-3 and contains approximately 34% pf ammonhia by mass(a) calculatew the mass of ammonia present in 1dm3 of the solution of this commercial ammonia(B)calculate the molarity of this solution (c) calculate the volume of distilled... 顯示更多 A commercial ammonia called "0.68 ammonia" has a density of 0.68 g com-3 and contains approximately 34% pf ammonhia by mass (a) calculatew the mass of ammonia present in 1dm3 of the solution of this commercial ammonia (B)calculate the molarity of this solution (c) calculate the volume of distilled water added if the solution has to be diluted to1 mol dm-4
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最佳解答:(a) 1dm3 = 1000cm3 1dm3 of the commercial ammonia weighs 0.68gcm-3 x 1000cm3 = 680 g Since the commercial ammonia contains 34% of ammonia by mass, the mass of the ammonia = 680 x 34% = 231.2 g (b) The RAM of the ammonia = 14 + 1 x 4 = 18 No. of mole of ammonia = 231.2 / 18 = 12.8 moles Molarity of ammonia = 12.8 moles / 1 dm3 = 12.8 M (c) Let a be the volume of water added. Hence 12.8 / (a + 1) = 1 12.8 = a +1 a = 11.8 dm3 Therefore, the volume of water added is 11.8 dm3. 2008-07-22 20:58:24 補充： (b) 應該係 14 + 1 x3 = 17 之後 231.2 / 17 = 13.6 13.6 / 1 = 13.6 M至啱 (c) 13.6 / a + 1 = 1 ，答案係 12.6 dm3
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