氯化钡物质的量:20.8 * 10% / 208 = 0.01mol
1、设硫酸钠x g
10x / (85.5 + x) * 142 = 0.01
解得:x = 14.15g
纯度:14.15 / 15 = 94.33%
2、溶质:0.02mol氯化钠
溶质质量:0.02 * 58.5 = 1.17g
溶液质量:10 + 20.8 - 233 * 0.01 = 28.47
质量分数:1.17 / 28.47 = 4.1%
氯化钡物质的量:20.8 * 10% / 208 = 0.01mol
1、设硫酸钠x g
10x / (85.5 + x) * 142 = 0.01
解得:x = 14.15g
纯度:14.15 / 15 = 94.33%
2、溶质:0.02mol氯化钠
溶质质量:0.02 * 58.5 = 1.17g
溶液质量:10 + 20.8 - 233 * 0.01 = 28.47
质量分数:1.17 / 28.47 = 4.1%