Final answer:
To prepare a 0.75 N solution of Al(NO₃)₃, 239.625 grams of Al(NO₃)₃ would be required to make 4.5 L of solution, considering aluminum's valency of 3 and the molar mass of Al(NO₃)₃.
Step-by-step explanation:
To calculate the mass of Al(NO₃)₃ needed to make a 0.75 N solution, we need to consider the normality equation, which is N = M x eqv, where 'N' is normality, 'M' is molarity, and 'eqv' is the number of equivalents. Since aluminum has a valency of 3, one mole of Al(NO₃)₃ will have 3 equivalents. Now, we need to rearrange the formula to find molarity: M = N / eqv, which gives M = 0.75 N / 3 eqv = 0.25 M.
Next, we calculate the number of moles of Al(NO₃)₃ required using the equation moles = volume x molarity: moles = 4.5 L x 0.25 M = 1.125 moles. Finally, we find the mass by multiplying the number of moles by the molar mass of Al(NO₃)₃ (molar mass of Al(NO₃)₃ is approximately 213 g/mol): mass = 1.125 moles x 213 g/mol = 239.625 grams of Al(NO₃)₃.