When 100 " mL of " 0.06 M Fe `(NO_3)_3,50mL ` of `0.2M FeCl_3` and `100mL` of `0.26M` Mg `(NO_3)_2`, are mixed In the final solution…..
`[Fe^(3+)]=` ….
`[NO_3^(ɵ)]=`…..
`[Cl^(ɵ)]=`……
`[Mg^(2+)]=`……

Explanation

To solve the problem, we need to calculate the concentrations of \( [Fe^{3+}] \), \( [NO_3^{-}] \), \( [Cl^{-}] \), and \( [Mg^{2+}] \) in the final solution after mixing the given solutions.

### Step-by-Step Solution:

1. **Calculate the total volume of the final solution:**
\[ \text{Total Volume} = 100 \, \text{mL} + 50 \, \text{mL} + 100 \, \text{mL} = 250 \, \text{mL} \]
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