If `m_(A)` gram of a metal A displaces `m_(B)` gram of another metal B from its salt solution and if the equilvalent mass are `E_(A) and E_(B)` respectively then equivalent mass of A can be expressed as:

Options

A. `E_(A)=(m_(A))/(m_(B)) xx E_(B)`
B. `E_(A)=(m_(A) xx m_(B))/(E_(B)`
C. `E_(A)= (m_(B))/(E_(A)) xx E_(B) `
D. `E_(A)= sqrt((m_(A))/(m_(B)) xx E_(B) )`

Answer

To express the equivalent mass of metal A in terms of the equivalent mass of metal B, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Reaction**:
Metal A displaces metal B from its salt solution. This means that metal A replaces metal B in a compound, typically represented as \( BSO_4 \) (sulfate of B), forming \( ASO_4 \) (sulfate of A).

2. **Defining Moles**:
Let \( m_A \) be the mass of metal A and \( m_B \) be the mass of metal B. The number of moles of metal A can be expressed as:
\[ n_A = \frac{m_A}{M_A} \]
where \( M_A \) is the molar mass of metal A.

Similarly, the number of moles of metal B is:
\[ n_B = \frac{m_B}{M_B} \]
where \( M_B \) is the molar mass of metal B.

3. **Equating Moles**:
Since one mole of A displaces one mole of B, we can set the number of moles equal to each other:
\[ n_A = n_B \]
Therefore,
\[ \frac{m_A}{M_A} = \frac{m_B}{M_B} \]

4. **Finding Equivalent Mass**:
The equivalent mass of a metal is defined as:
\[ E_A = \frac{M_A}{n_A} \quad \text{and} \quad E_B = \frac{M_B}{n_B} \]
In this case, if we assume the n-factor (valency) for both metals is 2 (as they are both in +2 oxidation state), we can express the equivalent masses as:
\[ E_A = \frac{M_A}{2} \quad \text{and} \quad E_B = \frac{M_B}{2} \]

5. **Relating Equivalent Masses**:
From the earlier equation \( \frac{m_A}{M_A} = \frac{m_B}{M_B} \), we can rearrange it to find \( M_A \):
\[ M_A = \frac{m_A \cdot M_B}{m_B} \]
Substituting this into the expression for \( E_A \):
\[ E_A = \frac{M_A}{2} = \frac{1}{2} \cdot \frac{m_A \cdot M_B}{m_B} \]

6. **Final Expression**:
Now, substituting \( M_A \) into the equivalent mass equation gives:
\[ E_A = E_B \cdot \frac{m_A}{m_B} \]
Therefore, the equivalent mass of A can be expressed in terms of the equivalent mass of B as:
\[ E_A = E_B \cdot \frac{m_A}{m_B} \]