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發問:
a water solution containing 100g ethanol, CH3CH3OH, per litre of solution has a density of 0.984 cm-3. Calculate: 1. the molarity of CH3CH3OH 2. the molarity of CH3CH3OH in this solution, and 3. the mole fraction of CH3CH3OH in this solution. 更新: answer for 1 is 2.171M, 2. is 2.46 molal, 3 is 0.042 更新 2: where does the 116 come from? 更新 3: i dont know the difference b/w Q1 and Q2, so i want to find out the answer in here.
最佳解答:
Taking 1 litre of such solution, then there are 984 g in which 100 g are ethanol and 884 g are water. So for (1), we have: 116 cm3 of ethanol weighs 100 g (Since vol. of water = 884 cm3) No. of moles of ethanol = 100/46 = 2.174 Hence molarity of ethanol = 2.174/0.116 = 18.74 M (2) Molarity of ethanol in this solution = 2.174 M (3) No. of moles of water = 884/18 = 49.1 So mole fraction of ethanol = 2.174/(2.174 + 49.1) = 0.0424 2009-03-20 17:39:41 補充： 116 comes from 1000 - 884 where 884 is the volume of water. In fact what's the dirrerence between Q1 and Q2 ?
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molarity calculation發問:
a water solution containing 100g ethanol, CH3CH3OH, per litre of solution has a density of 0.984 cm-3. Calculate: 1. the molarity of CH3CH3OH 2. the molarity of CH3CH3OH in this solution, and 3. the mole fraction of CH3CH3OH in this solution. 更新: answer for 1 is 2.171M, 2. is 2.46 molal, 3 is 0.042 更新 2: where does the 116 come from? 更新 3: i dont know the difference b/w Q1 and Q2, so i want to find out the answer in here.
最佳解答:
Taking 1 litre of such solution, then there are 984 g in which 100 g are ethanol and 884 g are water. So for (1), we have: 116 cm3 of ethanol weighs 100 g (Since vol. of water = 884 cm3) No. of moles of ethanol = 100/46 = 2.174 Hence molarity of ethanol = 2.174/0.116 = 18.74 M (2) Molarity of ethanol in this solution = 2.174 M (3) No. of moles of water = 884/18 = 49.1 So mole fraction of ethanol = 2.174/(2.174 + 49.1) = 0.0424 2009-03-20 17:39:41 補充： 116 comes from 1000 - 884 where 884 is the volume of water. In fact what's the dirrerence between Q1 and Q2 ?
其他解答:
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