When 0.054 moles of aluminum sulfate are dissolved in enough water to make 459 milliliters of solution, what is the molarity of sulfate ions? Answer in units of M.

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When 0.054 moles of aluminum sulfate are dissolved in enough water to make 459 milliliters of solution, what is the molarity of sulfate ions? Answer in units of M.

Chemistry
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Molarity = moles of solute (aluminum sulfate) divided by the volume of the solution in L. So, to calculate the molarity, divide the moles of aluminum sulfate by the volume of solution in L.
\[\sf \frac{0.054 ~mol~Al_2(SO_4)_3}{459~mL~soln}~\times ~ \frac{1~mL}{10^{-3}~L~ soln}~\times \frac{3~mol~SO_4^{2+}~ions}{1~mol~Al_2(SO_4)_3}\]\[\sf=\frac{(0.054 \cdot 3)~ mol~ SO_4^{2+} ~ions}{(459 \cdot 10^{-3}) ~L~ soln} \]\[=~?\]
.4 or .35 or .353 ??? depending on what you used for sig figs? @Jhannybean

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The answer I got was: \[\sf \frac{0.35 ~mol~ SO_$^{2+}~ions }{1~L~soln}\] 2 sig figs because you started with 0.054 moles aluminum sulfate
thanks so much!!!!

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