论文部分内容阅读
在复习关于溶解度与百分比浓度的混合计算时,往往涉及到结晶水合物的溶解度计算。这类题难度较大,学生易做错。例如: 已知无水硫酸铜溶解度0℃时为14.8克,40℃时为29克,把40℃时浓度为18％的硫酸铜溶液100克冷却到0℃,析出硫酸铜晶体多少克? 学生往往这样做: 解:溶质质量:100×18％=18(克) 溶剂质量:100-18=82(克) 设0℃时82克水应溶x克CuSO_4才饱和则100:14.8=82:x x=12.14(克) 相当于析出CuSO_4质量:18-12.14=5.86(克) 根据CuSO_4→CuSO_4·5H_2O,设析出晶
When reviewing the hybrid calculation of solubility and percent concentration, solubility calculations for crystalline hydrates are often involved. This type of problem is more difficult and students are more likely to make mistakes. For example: It is known that the solubility of anhydrous copper sulfate is 14.8 g at 0°C and 29 g at 40°C. After cooling 100 g of a 18% copper sulfate solution at 40°C to 0°C, how many grams of copper sulfate crystals will precipitate? This is often done: Solution: solute mass: 100 × 18% = 18 (g) Solvent mass: 100-18 = 82 (g) Set at 0 °C, 82 g of water should be dissolved x g CuSO 4 before saturation 100:14.8 = 82: Xx=12.14 (g) Equivalent to precipitation CuSO 4 mass: 18-12.14=5.86 (g) According to CuSO 4 → CuSO 4 · 5H 2 O, precipitated crystal