Answer:
[tex]m_{water}=1.708g[/tex]
Explanation:
Hello!
In this case, since the total mass of the sample is 2.714 g and the mass of the anhydrous salt is 1.006 g; we infer the mass of water is:
[tex]m_{water}=2.714g-1.006g\\\\m_{water}=1.708g[/tex]
Moreover, the percentage of water in the hydrate is calculated by dividing the mass of water by the mass of the total hydrate:
[tex]\% water=\frac{1.708g}{2.714g}*100\% \\\\\% water=62.9\%[/tex]
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