Question #041b5

1 Answer
Apr 20, 2017

Answer:

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.