初中《化学》第四章中的有关溶解度的计算是该章的一个重点,也是一个难点。对中差生要求完全掌握就更难了,笔者在几年的教学中坚持从概念出发,进行有关溶解度计算的教学,收到了较好的效果。特别对中差生学习和掌握这部分内容更为有效,现举例说明。一、温度不变时 [例1]已知20℃吋食盐的溶解度是36克,问20℃,150克的食盐饱和溶液里溶有 The calculation of the solubility in the fourth chapter of “Chemistry” in junior high school is a key point of this chapter, and it is also a difficult point. It is all the more difficult to master the requirements of the poor students. The author insisted on starting from the concepts in the teaching of several years, and carried out teaching on solubility calculation, and received good results. It is particularly effective for poor middle school students to learn and master this part of the content. First, when the temperature is constant [Example 1] It is known that the solubility of 20 °C forged salt is 36g, and it is dissolved in a saturated solution of 150g of salt at 20°C.