Volume `V_(1)mL` of 0.1 M `K_(2)Cr_(2)O_(7)` is needed for complete oxidation of 0.678g `N_(2)H_(4)` in acidic medium. The volume of 0.3M `KMnO_(4)` needed for same oxidation in acidic medium will be :

Options

A. `2/5V_1`
B. `5/2V_1`
C. `113 V_1`
D. can not be determined

Explanation

To solve the problem, we need to find the volume of 0.3 M KMnO4 required to completely oxidize 0.678 g of N2H4, given that a certain volume V1 mL of 0.1 M K2Cr2O7 is needed for the same reaction.

### Step-by-Step Solution:

1. **Calculate the number of moles of N2H4:**
- The molar mass of N2H4 (Hydrazine) is calculated as follows:
\[ \text{Molar mass of } N2H4 = (2 \times 14) + (4 \times 1) = 28 + 4 = 32 \text{ g/mol} \]
- Now, calculate the moles of N2H4 in 0.678 g:
\[ \text{Moles of } N2H4 = \frac{0.678 \text{ g}}{32 \text{ g/mol}} = 0.0211875 \text{ mol} \]

2. **Determine the equivalent of N2H4:**
- The reaction of N2H4 with K2Cr2O7 or KMnO4 requires knowing the number of electrons transferred. For N2H4, it is oxidized to N2, which involves a change of 4 electrons.
- Therefore, the equivalents of N2H4 can be calculated as:
\[ \text{Equivalents of } N2H4 = \text{Moles of } N2H4 \times 4 = 0.0211875 \text{ mol} \times 4 = 0.08475 \text{ eq} \]

3. **Calculate the equivalents of K2Cr2O7:**
- The n-factor for K2Cr2O7 in acidic medium is 6 (as derived from the oxidation state of Cr changing from +6 to +3).
- The equivalents of K2Cr2O7 can be calculated using the formula:
\[ \text{Equivalents of } K2Cr2O7 = \text{Molarity} \times \text{Volume (L)} \times n \]
- Given that the molarity is 0.1 M and volume is V1 mL (or V1/1000 L):
\[ \text{Equivalents of } K2Cr2O7 = 0.1 \times \frac{V1}{1000} \times 6 \]

4. **Set the equivalents equal:**
- Since the equivalents of N2H4 and K2Cr2O7 are equal:
\[ 0.08475 = 0.1 \times \frac{V1}{1000} \times 6 \]
- Simplifying gives:
\[ V1 = \frac{0.08475 \times 1000}{0.1 \times 6} = \frac{84.75}{0.6} = 141.25 \text{ mL} \]

5. **Calculate the equivalents of KMnO4:**
- The n-factor for KMnO4 in acidic medium is 5 (Mn changes from +7 to +2).
- The equivalents of KMnO4 can be calculated as:
\[ \text{Equivalents of } KMnO4 = \text{Molarity} \times \text{Volume (L)} \times n \]
- Let the volume of KMnO4 be V2 mL (or V2/1000 L):
\[ \text{Equivalents of } KMnO4 = 0.3 \times \frac{V2}{1000} \times 5 \]

6. **Set the equivalents equal:**
- Since the equivalents of N2H4 and KMnO4 are equal:
\[ 0.08475 = 0.3 \times \frac{V2}{1000} \times 5 \]
- Simplifying gives:
\[ V2 = \frac{0.08475 \times 1000}{0.3 \times 5} = \frac{84.75}{1.5} = 56.5 \text{ mL} \]

### Final Answer:
The volume of 0.3 M KMnO4 needed for the complete oxidation of 0.678 g of N2H4 in acidic medium is **56.5 mL**.