$m(\ce{FeSO4. $x$~H2O})=20.0\:\mathrm{g}$

$N(\ce{H2O})=3.03 \cdot10^{23}$


What I've done:


\begin{aligned} m (\ce{H2O})&= \frac{N(\ce{H2O}) \cdot M(\ce{H2O})}{N_{A}}\\ &= \frac{3.03 \cdot10^{23} \cdot 18.02$\:\mathrm{g/mol}}{6.02 \cdot 10^{23}/\:\mathrm{mol}}\\ &= 9.06987\:\mathrm{g} \end{aligned}


\begin{aligned} m(\ce{FeSO4})&= m(\ce{FeSO4} \cdot x \ce{H2O}) - m(\ce{H2O})\\ &= 20.00\:\mathrm{g} - 9.06987\:\mathrm{g}\\ &= 10.9301\:\mathrm{g} \end{aligned}

Any suggestions what to do next?


Calculate $n(\ce{FeSO4})$ with $M(\ce{FeSO4})\approx 136\:\mathrm{g/mol}$


Calculate $n(\ce{H2O})$


Set them into relation $n(\ce{FeSO4}):n(\ce{H2O})= 1:x$

$x\approx 6$


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