Answer:
[tex]\% H_2O=51.2\%[/tex]
Explanation:
Hello!
In this case, since the percent water is computed by dividing the amount of water by the total mass of the hydrate; we infer we first need the molar mass of water and that of the hydrate as shown below:
[tex]MM_{MgSO_4* 7H_2O}=120.36 g/mol+7*18.02g/mol\\\\MM_{MgSO_4* 7H_2O}=246.5g/mol[/tex]
Thus, the percent water is:
[tex]\% H_2O=\frac{7*MM_{H_2O}}{MM_{MgSO_4* 7H_2O}} *100\%\\\\[/tex]
So we plug in to obtain:
[tex]\% H_2O=\frac{7*18.02}{246.5} *100\%\\\\\% H_2O=51.2\%[/tex]
Best regards!
