Ex: 5 To 15 ml of 0.4mol/1 of Al_(2)(SO_(4))_(3) solution(S), 235 ml of water is added. A solution S_(1) is obtained. 1. Choose, from the following list, the appropriate material for this preparation. Available material list: * 5 mL, 10 mL and 20 mL volumetric pipets; * 5 mL, 10 mL and 20 mL graduated pipets; * 50 mL, 100 mL and 200 mL beakers; * 50 mL, 200 mL and 250 mL volumetric flasks; A pipet filler. 2. Calculate the molar concentration of the solution obtained. 3. Deduce the molar concentration of the ionic species presents in the solution S_(1) Aluminum sulfate solution

1. To prepare the solution, you would need a 20 mL volumetric flask to accurately measure and transfer the 15 mL of solution. Additionally, a pipet filler would be useful for measuring the exact volume of the solution.2. To calculate the molar concentration of the solution obtained, you can use the formula: where:- is the initial concentration of the solution (0.4 mol/L)- is the initial volume of the solution (15 mL)- is the final concentration of the solution (unknown)- is the final volume of the solution (15 mL + 235 mL = 250 mL)Rearranging the formula to solve for : Substituting the given values: Therefore, the molar concentration of the solution obtained is 0.024 mol/L.3. To deduce the molar concentration of the ionic species present in the solution , we need to consider the dissociation of in water. dissociates into 2 aluminum ions ( ) and 3 sulfate ions ( ) per formula unit. Therefore, the molar concentration of each ionic species will be:For : For : Therefore, the molar concentration of the ionic species present in the solution is 0.048 mol/L for and 0.072 mol/L for .