2. Some sample of aluminum chloride weighing 1.0000 g was dissolved in water and 250.0 ml of solution was obtained. To 10.00 ml of this solution, 20.00 ml of 0.025mol/1 EDTA solution was added. F_(0,025)=1.100 , and the resulting solution was titrated with eriochrome black T by 12.50 ml of 0.02500mol/L magnesium sulfate solution. Calculate the mass fraction of aluminum in the sample in %

To calculate the mass fraction of aluminum in the sample, we need to follow these steps:<br /><br />1. Calculate the concentration of EDTA in the 10.00 ml solution.<br />2. Calculate the concentration of magnesium ions in the 10.00 ml solution.<br />3. Calculate the concentration of aluminum ions in the 10.00 ml solution.<br />4. Calculate the mass fraction of aluminum in the original sample.<br /><br />Given information:<br />- Mass of aluminum chloride sample: 1.0000 g<br />- Volume of solution obtained: 250.0 ml<br />- Volume of 10.00 ml solution taken for titration: 10.00 ml<br />- Concentration of EDTA solution: 0.025 mol/L<br />- Titration factor (F0,025) for EDTA: 1.100<br />- Volume of magnesium sulfate solution used for titration: 12.50 ml<br />- Concentration of magnesium sulfate solution: 0.02500 mol/L<br /><br />Step 1: Calculate the concentration of EDTA in the 10.00 ml solution.<br />Concentration of EDTA in 10.00 ml solution = (0.025 mol/L) × (10.00 ml / 250.00 ml) = 0.001 mol/L<br /><br />Step 2: Calculate the concentration of magnesium ions in the 10.00 ml solution.<br />Concentration of magnesium ions in 10.00 ml solution = (0.02500 mol/L) × (12.50 ml / 250.00 ml) = 0.00125 mol/L<br /><br />Step 3: Calculate the concentration of aluminum ions in the 10.00 ml solution.<br />Using the titration factor (F0,025) for EDTA, we can calculate the concentration of aluminum ions in the 10.00 ml solution.<br />Concentration of aluminum ions in 10.00 ml solution = (0.001 mol/L) × (1.100) = 0.0011 mol/L<br /><br />Step 4: Calculate the mass fraction of aluminum in the original sample.<br />Mass of aluminum in the 10.00 ml solution = (0.0011 mol/L) × (10.00 ml) × (27.00 g/mol) = 0.0297 g<br />Mass fraction of aluminum in the original sample = (0.0297 g / 1.0000 g) × 100% = 2.97%<br /><br />Therefore, the mass fraction of aluminum in the sample is 2.97%.