Question

Calculate the mass (in g) of anhydrous aluminum sulfate (MM = 342.15 g/mol) needed to make 250.0 mL of a 0.850 M Al3+ solution.

Answer 1

The mass of anhydrous aluminum sulfate needed will be 73.48 grams.

Stoichiometric problem

Recall that:

Molarity = number of moles of solute/volume of solution

Also,

Mole = mass/molar mass.

Thus,

0.859 M = mole/0.250

mole = 0.21475 moles of aluminum sulfate.

Mass of 0.21475 moles = mole x molar mass

                                  = 0.21475 x 342.15

                                       = 73.48 grams

Thus, the mass of anhydrous aluminum sulfate needed to make 250.0 mL of 0.850 M solution will be 73.48 grams.

More on stoichiometric problems can be found here: brainly.com/question/28297916

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