57.称取1.032g氧化铝试样,溶解定容至250mL容量瓶中,移取25.00mL,加入TA12O3/EDTA=1.505mgl/mL的EDTA涪液10.00mL,以二甲酚橙为指示剂,用Zn( )2标准滴定溶液12.20ml,滴定至终点,已知20.00mLZn( )2溶液相当于13.62ml.EDTA溶液。则试样中Al2O3的含量为()。

题目解析 In this question, we are determining the content of aluminum oxide (Al2O3) in a given sample. The sample weighs 1.032g and is dissolved in water to make a 250mL solution. A 25.00mL portion of this solution is titrated with a standard zinc (Zn)2+ solution. The titration is performed using 10.00mL of EDTA reagent (concentration 1.505mgl/mL) with xylenol orange as the indicator. It is given that 20.00mL of the Zn2+ solution is equivalent to 13.62mL of the EDTA solution. The balanced equation for the titration reaction is: Zn(EDTA)2- + Al2O3 → Al(EDTA)2- + Zn2+ First, we need to find the number of moles of EDTA used in the titration. Number of moles of EDTA = concentration × volume Number of moles of EDTA = 1.505mgl/mL × 0.01000L = 0.01505mmol From the equation, we can see that 1 mole of EDTA reacts with 1 mole of Al2O3. However, the Zn2+ solution is given in terms of equivalence with the EDTA solution. Therefore, we need to find the number of moles of Al2O3 using the equivalence. Number of moles of Al2O3 = (equivalence volume of Zn2+ solution × number of moles of EDTA) / volume of Al2O3 sample Number of moles of Al2O3 = (12.20mL × 0.01505mmol) / 25.00mL = 0.007302mmol Mass of Al2O3 in the sample = number of moles × molar mass Mass of Al2O3 in the sample = 0.007302mmol × 101.96g/mol ≈ 0.7443g Percentage of Al2O3 in the sample = (mass of Al2O3 / mass of the sample) × 100 Percentage of Al2O3 in the sample = (0.7443g / 1.032g) × 100 ≈ 72.21% Therefore, the correct answer is D. 72.21%.