To calculate the volume of water needed to create a 0.450 M solution of Al(NO3)3, we need to use the formula:
Molarity = moles of solute / volume of solution in liters
First, we need to determine the number of moles of Al(NO3)3 we have:
moles of Al(NO3)3 = mass / molar mass
molar mass of Al(NO3)3 = 1 x atomic mass of Al + 3 x atomic mass of N + 9 x atomic mass of O = 1 x 26.98 + 3 x 14.01 + 9 x 16.00 = 212.99 g/mol
moles of Al(NO3)3 = 41.6 g / 212.99 g/mol = 0.195 mol
Next, we can rearrange the formula above to solve for the volume of solution:
volume of solution = moles of solute / molarity
volume of solution = 0.195 mol / 0.450 M = 0.433 L
Therefore, we need to add 0.433 L (or 433 mL) of water to 41.6 g of Al(NO3)3 to create a 0.450 M solution.