# Question #041b5

Apr 20, 2017

0.25 g of anhydrous $A l C {l}_{3}$.

#### Explanation:

Use the relationship $M = \frac{n}{V}$ 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 $\left(\frac{125}{1000}\right) =$0.001875 mol.

Then multply this number by the molar mass of $A l C {l}_{3}$ which is 133.341 g/mol based on the anhydrous form.

0.001875 x 133.341 = 0.25 g.