称取干燥好的工业纯碱试样1.5432g,加水溶解后转入250ml容量瓶中定容.移取此试液25.00ml,以甲基橙为指示剂,用0.1000HCl标准溶液滴定至终点,消耗HCl 24.68ml求试样中Na2CO3的含量.

现计算出你配的溶液的摩尔浓度,n=m/M=1.5432g/106g/mol=0.0146mol,c=n/v=0.0146/0.25mol/l=0.0584mol/l,实际取出25ml,n=cv1=0.0584*0.025mol=0.00146mol
消耗的盐酸,n=c*v=0.1000*0.02468mol=0.002468mol,
盐酸和碳酸钠反应消耗比例是2：1,
那么参与反映的碳酸钠是0.002468/2mol=0.001234mol
含量,0.001234/0.00146=0.8452=85.2%

记得采纳