How many gram atoms of oxygen are contained in 0.25 mole of `Na_(2)S_(2)O_(3).5H_(2)O` (hypo)?

Options

A. `0.5`
B. `0.75`
C. `2.0`
D. `1.20`

Explanation

To find out how many gram atoms of oxygen are contained in 0.25 mole of `Na₂S₂O₃·5H₂O` (hypo), we can follow these steps:

### Step 1: Determine the number of oxygen atoms in one formula unit of `Na₂S₂O₃·5H₂O`.

In one formula unit of `Na₂S₂O₃·5H₂O`, we can break it down as follows:
- The `Na₂S₂O₃` part contains 3 oxygen atoms.
- The `5H₂O` part contains 5 water molecules, and each water molecule has 1 oxygen atom. Therefore, 5 water molecules contribute 5 oxygen atoms.

Now, we can calculate the total number of oxygen atoms:
- Total oxygen atoms = Oxygen from `Na₂S₂O₃` + Oxygen from `5H₂O`
- Total oxygen atoms = 3 + 5 = 8 oxygen atoms.

### Step 2: Calculate the number of gram atoms of oxygen in 0.25 moles of `Na₂S₂O₃·5H₂O`.

Since 1 mole of `Na₂S₂O₃·5H₂O` contains 8 gram atoms of oxygen, we can find the number of gram atoms of oxygen in 0.25 moles:
- Number of gram atoms of oxygen = Number of gram atoms in 1 mole × Number of moles
- Number of gram atoms of oxygen = 8 × 0.25 = 2 gram atoms of oxygen.

### Conclusion

Thus, in 0.25 mole of `Na₂S₂O₃·5H₂O`, there are **2 gram atoms of oxygen**.

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