"Equivalent mass" =("Molecular mass/Atomic mass" )/("n-factor")
n-factor is very important in redox as well as non-redox reactions.With the help of n-factor we can predict the molar ratio of the reactant species specis taking part in reactions. The reciprocal of n-factor's ratio of the reactions is the molar ratio of the reactants.
In general n-factor of acid/base is number of moles of `H^(+)//OH^(-)` furnished per mole of acid/base n-factor of a reactant is number of moles electrons lost or gained per mole of reactant.
Example 1:
(1)In acidic medium : `KMnO_(4) (n=5)to Mn^(2+)`
(2) In neutral medium : `KMnO_(4)(n=3) to Mn^(2+)`
(3) In basic medium : `KMnO_(4)(n=1) to Mn^(6+)`
Example 2 : `FeC_(2)O_(4)to Fe^(3+)+2CO_(2)`
Total number of moles `e^(-)` lost by 1 mole of `FeC_(2)O_(4)`
`=1+1xx2 implies 3`
For the reaction, `O("molar mass=M") to Fe_(2)O_(3)` what is the eq. mass of `fe_(0.95)`O ?

Options

A. `(M)/(0.85)`
B. `(M)/(0.95)`
C. `(M)/(0.8075)`
D. None of these

Answer

To solve the problem of finding the equivalent mass of Fe0.95O, we will follow these steps:

### Step 1: Determine the Molar Mass of Fe0.95O
The molar mass of Fe0.95O can be calculated using the atomic masses of iron (Fe) and oxygen (O). The atomic mass of Fe is approximately 55.85 g/mol and that of O is approximately 16.00 g/mol.

\[ \text{Molar mass of Fe}_{0.95}\text{O} = (0.95 \times 55.85) + (1 \times 16.00) \]

Calculating this gives:

\[ \text{Molar mass of Fe}_{0.95}\text{O} = (53.0575) + (16.00) = 69.0575 \, \text{g/mol} \]

### Step 2: Determine the Oxidation State of Iron in Fe0.95O
To find the n-factor, we need to determine the change in oxidation state of iron in the reaction from Fe0.95O to Fe2O3.

Let the oxidation state of Fe in Fe0.95O be \( x \). The formula for Fe0.95O can be expressed as:

\[ 0.95x + (-2) = 0 \quad \text{(since O has an oxidation state of -2)} \]

Solving for \( x \):

\[ 0.95x = 2 \implies x = \frac{2}{0.95} \approx 2.105 \]

### Step 3: Determine the Change in Oxidation State
In Fe2O3, the oxidation state of Fe is +3. Therefore, the change in oxidation state when Fe0.95O is converted to Fe2O3 is:

\[ \text{Change in oxidation state} = 3 - 2.105 = 0.895 \]

### Step 4: Calculate the n-factor
The n-factor is calculated as the change in oxidation state multiplied by the number of moles of Fe in Fe0.95O:

\[ \text{n-factor} = 0.895 \times 0.95 \approx 0.85025 \approx 0.85 \]

### Step 5: Calculate the Equivalent Mass
The equivalent mass of Fe0.95O can now be calculated using the formula:

\[ \text{Equivalent mass} = \frac{\text{Molar mass}}{\text{n-factor}} = \frac{69.0575}{0.85} \]

Calculating this gives:

\[ \text{Equivalent mass} \approx 81.23 \, \text{g/equiv} \]

### Final Answer
The equivalent mass of Fe0.95O is approximately **81.23 g/equiv**.

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