Question

Question #041b5

Answer 1

0.25 g of anhydrous `AlCl_3`.

Explanation:

Use the relationship `M = n/V` where M is the molarity in moles per litre, n is the number of moles, and V is the volume in litres. Rearrange for n to give `n = M.V` and plug in the numbers:

`n = M.V = 0.0150` x `(125/1000) =`0.001875 mol.

Then multply this number by the molar mass of `AlCl_3` which is 133.341 g/mol based on the anhydrous form.

0.001875 x 133.341 = 0.25 g.