In the titration of `K_(2)Cr_(2)O_(7)` and ferrous sulphate, following data is obtained:
`V_(1)` mL of `K_(2)Cr_(2)O_(7)` solution of molarity `M_(1)` requires `V_(2)`mL of `FeSO_(4)` solution of molarity `M_(2)`.
Which of the following relations is /are true for the above titration?

Options

A. `6M_(1)V_(1)=M_(2)V_(2)`
B. `M_(1)V_(1)=6M_(2)V_(2)`
C. `N_(1)V_(1)=N_(2)V_(2)`
D. `M_(1)V_(1)=M_(2)V_(2)`

Answer

To solve the problem regarding the titration of \( K_2Cr_2O_7 \) and ferrous sulfate (\( FeSO_4 \)), we will derive the relationships based on the equivalence of the two reactants involved in the reaction.

### Step-by-Step Solution:

1. **Identify the Reaction**:
The balanced chemical reaction between potassium dichromate (\( K_2Cr_2O_7 \)) and ferrous sulfate (\( FeSO_4 \)) can be represented as:
\[ K_2Cr_2O_7 + 6FeSO_4 + 7H_2SO_4 \rightarrow 2Cr_2(SO_4)_3 + 3Fe_2(SO_4)_3 + K_2SO_4 + 7H_2O \]
From this equation, we can see that 1 mole of \( K_2Cr_2O_7 \) reacts with 6 moles of \( FeSO_4 \).

2. **Determine the n-factor**:
- For \( K_2Cr_2O_7 \): The oxidation state of chromium changes from +6 to +3. The change in oxidation state is 3 for each chromium atom, and there are 2 chromium atoms. Therefore, the n-factor for \( K_2Cr_2O_7 \) is \( 2 \times 3 = 6 \).
- For \( FeSO_4 \): The oxidation state of iron changes from +2 to +3. The change is 1 for each iron atom, and there is 1 iron atom. Thus, the n-factor for \( FeSO_4 \) is 1.

3. **Apply the Law of Equivalence**:
The law of equivalence states that the equivalents of the reactants must be equal at the equivalence point:
\[ \text{Equivalents of } K_2Cr_2O_7 = \text{Equivalents of } FeSO_4 \]

4. **Calculate Equivalents**:
The equivalents can be calculated using the formula:
\[ \text{Equivalents} = \text{Molarity} \times \text{Volume} \times \text{n-factor} \]
For \( K_2Cr_2O_7 \):
\[ \text{Equivalents of } K_2Cr_2O_7 = M_1 \times V_1 \times 6 \]
For \( FeSO_4 \):
\[ \text{Equivalents of } FeSO_4 = M_2 \times V_2 \times 1 \]

5. **Set Equivalents Equal**:
Setting the equivalents equal gives:
\[ M_1 \times V_1 \times 6 = M_2 \times V_2 \]
This is one of the relationships derived from the titration.

6. **Normality Relation**:
We can also express the relationship in terms of normality. Normality (N) is related to molarity (M) and n-factor as follows:
\[ N = M \times n \]
Therefore, for \( K_2Cr_2O_7 \) and \( FeSO_4 \):
\[ N_1 = M_1 \times 6 \quad \text{and} \quad N_2 = M_2 \times 1 \]
The equivalents can also be expressed as:
\[ N_1 \times V_1 = N_2 \times V_2 \]

### Final Relationships:
Thus, the relationships that hold true for the titration are:
1. \( 6M_1V_1 = M_2V_2 \)
2. \( N_1V_1 = N_2V_2 \)