1.420 g of anhydrous `ZnSO_(4)` was left in moist air. After a few days its weight was found to be 2.528 g. How many water molecules are present in its hydrated salt formula (molar mass of `ZnSO_(4)` = 161.5)?

Explanation

To find out how many water molecules are present in the hydrated salt formula of zinc sulfate (ZnSO₄), we will follow these steps:

### Step 1: Calculate the mass of water absorbed
The initial mass of anhydrous ZnSO₄ is given as 1.420 g, and after exposure to moist air, the mass increased to 2.528 g.

To find the mass of water absorbed, we subtract the initial mass from the final mass:

\[ \text{Mass of water absorbed} = \text{Final mass} - \text{Initial mass} = 2.528 \, \text{g} - 1.420 \, \text{g} = 1.108 \, \text{g} \]

### Step 2: Calculate the number of moles of anhydrous ZnSO₄
Using the molar mass of ZnSO₄, which is 161.5 g/mol, we can calculate the number of moles of anhydrous ZnSO₄:

\[ \text{Moles of ZnSO₄} = \frac{\text{Mass of ZnSO₄}}{\text{Molar mass of ZnSO₄}} = \frac{1.420 \, \text{g}}{161.5 \, \text{g/mol}} \approx 0.00879 \, \text{mol} \]

### Step 3: Calculate the number of moles of water absorbed
Next, we need to find the number of moles of water that corresponds to the mass of water absorbed. The molar mass of water (H₂O) is approximately 18 g/mol.

\[ \text{Moles of water} = \frac{\text{Mass of water}}{\text{Molar mass of water}} = \frac{1.108 \, \text{g}}{18 \, \text{g/mol}} \approx 0.06156 \, \text{mol} \]

### Step 4: Determine the ratio of moles of water to moles of ZnSO₄
Now, we can find the ratio of moles of water to moles of ZnSO₄ to determine how many water molecules are associated with each formula unit of ZnSO₄:

\[ \text{Ratio} = \frac{\text{Moles of water}}{\text{Moles of ZnSO₄}} = \frac{0.06156 \, \text{mol}}{0.00879 \, \text{mol}} \approx 7 \]

### Conclusion
The ratio indicates that there are approximately 7 water molecules associated with each formula unit of ZnSO₄ in the hydrated salt.

Thus, the hydrated salt formula is \( \text{ZnSO}_4 \cdot 7\text{H}_2\text{O} \).