Experimentally it was found that a metal oxide in formula `M_(0.98)O`. Metal `M` is present as `M^(2+)` and `M^(3+)` in its oxide ,Fraction of the metal which exists as `M^(3+)` would be

Answer

To solve the problem of finding the fraction of the metal M that exists as \( M^{3+} \) in the oxide \( M_{0.98}O \), we can follow these steps:

### Step 1: Understand the composition of the metal oxide
The formula \( M_{0.98}O \) indicates that there are 0.98 moles of metal M for every mole of oxygen. We can assume that the total mass of oxygen is 1.0 g, which means the mass of metal M is 0.98 g.

### Step 2: Define the oxidation states
We know that the metal M exists in two oxidation states: \( M^{2+} \) and \( M^{3+} \). Let:
- \( x \) = moles of \( M^{2+} \)
- \( 0.98 - x \) = moles of \( M^{3+} \)

### Step 3: Write the charge balance equation
Since the compound is neutral, the total positive charge must equal the total negative charge from the oxygen. The charge balance can be expressed as:
\[ 2x + 3(0.98 - x) - 2 = 0 \]
Here, the charge from oxygen is \( -2 \) (since there is one oxygen atom contributing a \( -2 \) charge).

### Step 4: Simplify the equation
Expanding the equation gives:
\[ 2x + 2.94 - 3x - 2 = 0 \]
Combining like terms results in:
\[ - x + 0.94 = 0 \]

### Step 5: Solve for \( x \)
Rearranging gives:
\[ x = 0.94 \]
This means there are 0.94 moles of \( M^{2+} \).

### Step 6: Find the moles of \( M^{3+} \)
Now we can find the moles of \( M^{3+} \):
\[ 0.98 - x = 0.98 - 0.94 = 0.04 \]
Thus, there are 0.04 moles of \( M^{3+} \).

### Step 7: Calculate the fraction of \( M^{3+} \)
The fraction of metal M that exists as \( M^{3+} \) is given by:
\[ \text{Fraction of } M^{3+} = \frac{\text{moles of } M^{3+}}{\text{total moles of M}} = \frac{0.04}{0.98} \]
Calculating this gives:
\[ \text{Fraction of } M^{3+} = 0.0408 \approx 0.041 \]

### Final Answer
The fraction of the metal M that exists as \( M^{3+} \) is approximately **0.041**.

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