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?

  • 1
    $\begingroup$ I have improved the formatting of your post using $\LaTeX$. For more information on how to do this yourself please see here and here. Please avoid using Latex in titles due to searching issues $\endgroup$
    – bon
    Commented Sep 3, 2015 at 16:07
  • $\begingroup$ Now it looks to me that you are right, and the given answer is wrong. $\endgroup$ Commented Sep 3, 2015 at 16:27

1 Answer 1


You're right and the given answer is wrong. Here's the math:

$$\left(1\;\mathrm{mol}\;\ce{Fe2(SO4)3}\over {399.88\;\mathrm{g}\;\ce{Fe2(SO4)3}}\right)\left({2.0\;\mathrm{g}\;\ce{Fe2(SO4)3}\over 100\;\mathrm{cm}^{3}}\right)\left({3\;\mathrm{mol}\;\ce{SO4^2-}\over 1\;\mathrm{mol}\;\ce{Fe2(SO4)3}}\right)\left({10\;\mathrm{cm}\over 1\;\mathrm{dm}}\right)^{3} =\ce{[SO4^2- ]} $$

$$\ce{[SO4^2- ]}= 1.5\cdot10^{-1}\;\mathrm{mol}\;\mathrm{dm}^{-3}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.