如图是配制溶质质量分数为10%的NaCl溶液的实验操作示意图:(1)用图1表示的序号表示配制溶液的正确操作

2025-03-15 14:57:49
推荐回答(1个)
回答1:

(1)根据配制溶液的步骤:称量食盐的质量,首先打开广口瓶塞,倒放在桌面上,将取得的食盐放在天平的左盘,然后将称得的食盐放到烧杯中,再将量取的水倒入烧杯中,用玻璃棒搅拌使之溶解,故正确的顺序为②⑤①④③;
(2)图②中的塑料仪器为药匙;
(3)食盐的质量=砝码+游码,据图可知,砝码的读数是15g,游码的读数是3.2g,故食盐的质量=15g+3.2g=18.2g;
(4))①用量筒量取水时仰视读数,将导致所取水体积偏大,导致溶质质量分数偏小;
②将量好的水倒入烧杯时有少量水溅出,将导致所取水体积偏小,导致溶质质量分数偏大;
③氯化钠晶体不纯,会使所得溶质质量中一部分为杂质质量,所得氯化钠质量偏小,导致溶质质量分数偏小;
④配制溶液的烧杯用少量蒸馏水润洗会使所取水质量增加,导致溶质质量分数偏小;
⑤溶液在烧杯配好后转移到试剂瓶中时有少量洒出,溶液具有均一性,所以洒落的溶液不会影响溶液的质量分数;
所以①③④正确;
故答案为:
(1)②⑤①④③;
(2)药匙;
(3)18.2g;
(4)①③④.

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