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

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


$x=?$

What I've done:

(1.)

\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}

(2.)

\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?

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

$n(\ce{FeSO4})\approx0.08\:\mathrm{mol}$

Calculate $n(\ce{H2O})$

$n(\ce{H2O})\approx0.5\:\mathrm{mol}$

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

$x\approx 6$

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