A sample of $\mathrm{2.00~g}$ of iron(III) sulphate is dissolved in water to give $\mathrm{100~cm^3}$ of aqueous solution. What is the concentration of $\ce{SO4^2-}$ ions in $\mathrm{mol~dm^{-3}}$?
My attempt:
I found the number of moles which is $\mathrm{0.005~mol}$.
Then, 1 mol of iron(III) sulphate has 3 mol of $\ce{SO4^2-}$
So the concentration is $\mathrm{\frac{0.005}{0.1} \cdot 3 = 1.5 \cdot 10^{-1} ~mol~dm^{-3}}$
But the given answer is $\mathrm{1.5 \cdot 10^{-2}~mol~dm^{-3}}$. Why?