Question #041b5

1 Answer
Simon Moore
Apr 20, 2017

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.

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