In a reaction, `Cr_(2)O_(7)^(2-)` is reduced to `Cr^(3+)`. What is concentration of `0.1M K_(2)Cr_(2)O_(7)` in equivalent per litre?
`Cr_(2)O_(7)^(2-)+14H^(+)+6erarr2Cr^(3+)+7H_(2)O`

Options

A. 0.9 N
B. 0.6 N
C. 0.3 N
D. 0. 2 N

Answer

To solve the problem, we need to determine the concentration of `0.1 M K2Cr2O7` in equivalents per liter. The relevant half-reaction provided is:

\[ \text{Cr}_2\text{O}_7^{2-} + 14 \text{H}^+ + 6 \text{e}^- \rightarrow 2 \text{Cr}^{3+} + 7 \text{H}_2\text{O} \]

### Step-by-Step Solution:

1. **Identify the Molarity of the Solution**:
The concentration of the potassium dichromate solution is given as `0.1 M`.

2. **Calculate the Number of Moles**:
Since the volume of the solution is 1 liter, we can calculate the number of moles of `K2Cr2O7`:
\[ \text{Number of moles} = \text{Molarity} \times \text{Volume} = 0.1 \, \text{mol/L} \times 1 \, \text{L} = 0.1 \, \text{mol} \]

3. **Determine the n-factor**:
The n-factor is defined as the number of electrons transferred in the reaction. From the half-reaction, we see that `6 electrons` are involved in the reduction of `Cr2O7^{2-}` to `Cr^{3+}`. Therefore, the n-factor is:
\[ n = 6 \]

4. **Calculate Normality**:
Normality (N) is calculated using the formula:
\[ \text{Normality} = \text{Molarity} \times n \]
Substituting the values we have:
\[ \text{Normality} = 0.1 \, \text{M} \times 6 = 0.6 \, \text{N} \]

5. **Conclude the Concentration in Equivalents per Liter**:
Since normality is equivalent to the concentration in equivalents per liter, we conclude that:
\[ \text{Concentration in equivalents per liter} = 0.6 \, \text{eq/L} \]

### Final Answer:
The concentration of `0.1 M K2Cr2O7` in equivalents per liter is **0.6 eq/L**.