想問幾條F.4Chemistry的問題~~ 關於計mole架 @ lgzrelv的部落格

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想問幾條F.4Chemistry的問題~~ 關於計mole架

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唉,,, 我完全都唔識mole呀...以下呢幾條chem我真係唔多識麻煩大家幫我拆拆佢唔該,,, 不勝感激!!Q1. You are given 25.0cm^3 of 18.0M hydrochloric acid. Caculate the molarities if the following volumes of water are added successively: 25.0 cm^3, 75.0 cm^3, 100 cm^3.Q2. A commercial ammonia called '0.68 ammonia' has a density of 0.68... 顯示更多唉,,, 我完全都唔識mole呀... 以下呢幾條chem我真係唔多識麻煩大家幫我拆拆佢唔該,,, 不勝感激!! Q1. You are given 25.0cm^3 of 18.0M hydrochloric acid. Caculate the molarities if the following volumes of water are added successively: 25.0 cm^3, 75.0 cm^3, 100 cm^3. Q2. A commercial ammonia called '0.68 ammonia' has a density of 0.68 g cm^-3 and contains approximately 34% of ammonia by mass. a). Caculate the mass of ammonia present in 1 dm^3 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 to 1 mol dm^-3. Q3. A student is going to prepare a 250 cm^3 solution of oxalic acid from its pure crytals ((COOH)2 x 2H2O). He weighs out 1.8g of oxalic acid crystals and dissolves them in 25cm^3 of distilled water. Then the solution is diluted to 250cm^3. a). Calculate the number of moles of oxalic acid abtained from 1.8g of its crystals. b). Calculate the molarity of the 25 cm^3 solution of oxalic acid. c). Calculate the molarity of the solution of oxalic acid after it has been diluted to 250 cm^3. 超級感謝高人幫我解答,,, 麻煩show 下個steps,,, 如過有D解釋就會prefect喇不過冇解釋都唔緊要,,, 最緊要就要有steps,,, 如果唔係我會嚴重看不懂...

最佳解答:

Q1. Original solution: M1 = 18 M. V1 = 25/1000 dm3 When 25 cm3 of water is added, V2 = (25 + 25)/1000 = 50/1000 dm3 M1V1 = M2V2 18(25/1000) = M2(50/1000) Hence, molarity M2 = 9 M When further 75 cm3 of water is added, V2 = (50 + 75)/1000 = 125/1000 dm3 M1V1 = M2V2 18(25/1000) = M2(125/1000) Hence, molarity M2 = 3.6 M When further 100 cm3 of water is added, V2 = (125 + 100)/1000 = 225/1000 dm3 M1V1 = M2V2 18(25/1000) = M2(225/1000) Hence, molarity M2 = 2 M Q2. a) Volume of the solution = 1 dm3 = 1000 cm3 Mass of the solution = mass x density = 0.68 x 1000 = 680 g Mass of NH3 = 680 x 34% = 231.2 g b) Molar mass of NH3 = 14 + 1x3 = 17 g mol-1 No. of moles of NH3 = mass/(molar mass) = 231.2/17 = 13.6 mol Molarity of NH3 = mol/volume = 13.6/1 = 13.6 M Q.3 a) Molar mass of (COOH)2?2H2O = 12x2 + 16x6 + 1x6 = 126 g mol-1 No. of moles of (COOH)2?2H2O = mass/molar mass = 1.8/126 = 0.0143 mol b) When volume = 25 cm3 = 0.025 dm3 Molarity = mol/volume = 0.0143/0.025 = 0.572 M c) When volume = 250 cm3 = 0.25 dm3 Molarity = 0.0143/0.25 = 0.0572 M =

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