56.称取氯化锌试样0.3600g,溶于水后控制溶液的酸度pH=6。以二甲酚橙为指示剂,用0.1024mol/L的EDTA溶液25.00mL滴定至终点,则氯化锌的含量为()。M(ZnCl2)=136.29g/mol。

题目解析 In this question, we are determining the content of zinc chloride (ZnCl2) in a given sample. The sample weighs 0.3600g and is dissolved in water to a solution with a pH of 6. The titration is performed using 0.1024mol/L EDTA solution (25.00mL) with xylenol orange as the indicator. The molar mass of ZnCl2 is given as 136.29g/mol. The balanced equation for the titration reaction is: ZnCl2 + EDTA → Zn(EDTA)2- + 2Cl- To calculate the content of ZnCl2, we need to determine the number of moles of EDTA used in the titration. The volume of EDTA solution used is 25.00mL. Number of moles of EDTA = concentration × volume Number of moles of EDTA = 0.1024mol/L × 0.02500L = 0.00256mol Since 1 mole of ZnCl2 reacts with 1 mole of EDTA, the number of moles of ZnCl2 in the sample is also 0.00256mol. Mass of ZnCl2 in the sample = number of moles × molar mass Mass of ZnCl2 in the sample = 0.00256mol × 136.29g/mol ≈ 0.3489g Percentage of ZnCl2 in the sample = (mass of ZnCl2 / mass of the sample) × 100 Percentage of ZnCl2 in the sample = (0.3489g / 0.3600g) × 100 ≈ 96.92% Therefore, the correct answer is A. 96.92%.