1. The problem involves determining the number of water molecules of hydration, $n$, in the compound $\text{MgSO}_4 \cdot n \text{H}_2\text{O}$. 2. Given data: 47.2% water content by mass, and the mole ratio $\frac{6}{120} = 0.05$ moles. 3. The chemical equation is: $$\text{MgSO}_4 \cdot n \text{H}_2\text{O} \rightarrow \text{MgSO}_4 + n \text{H}_2\text{O}$$ 4. To find $n$, use the formula for percent composition by mass: $$\% \text{water} = \frac{n \times M_{\text{H}_2\text{O}}}{M_{\text{MgSO}_4 \cdot n \text{H}_2\text{O}}} \times 100$$ where $M_{\text{H}_2\text{O}} = 18$ g/mol and $M_{\text{MgSO}_4} = 120$ g/mol. 5. The molar mass of the hydrate is: $$M_{\text{MgSO}_4 \cdot n \text{H}_2\text{O}} = 120 + 18n$$ 6. Substitute the given percent water (47.2%) into the equation: $$47.2 = \frac{18n}{120 + 18n} \times 100$$ 7. Simplify and solve for $n$: $$0.472 = \frac{18n}{120 + 18n}$$ Multiply both sides by $120 + 18n$: $$0.472(120 + 18n) = 18n$$ $$56.64 + 8.496n = 18n$$ $$56.64 = 18n - 8.496n = 9.504n$$ $$n = \frac{56.64}{9.504} = 5.96 \approx 6$$ 8. The calculated $n$ is approximately 6, indicating the hydrate is $\text{MgSO}_4 \cdot 6 \text{H}_2\text{O}$. 9. The given $n=7$ might be a rounding or experimental value, but calculation shows $n \approx 6$. Final answer: $n = 6$ water molecules of hydration.