Volume of 0.1 M `K_(2)Cr_(2)O_(7)` required to oxidize 35 mL of 0.5 M `FeSO_(4)` solution is

Options

A. 26.2 mL
B. 175 mL
C. 185 mL
D. 145 mL

Answer

To find the volume of 0.1 M \( K_2Cr_2O_7 \) required to oxidize 35 mL of 0.5 M \( FeSO_4 \) solution, we can follow these steps:

### Step 1: Determine the number of moles of \( FeSO_4 \)

First, we need to calculate the number of moles of \( FeSO_4 \) in the given volume.

\[ \text{Moles of } FeSO_4 = \text{Molarity} \times \text{Volume (in L)} \]

Given:
- Molarity of \( FeSO_4 = 0.5 \, M \)
- Volume of \( FeSO_4 = 35 \, mL = 0.035 \, L \)

\[ \text{Moles of } FeSO_4 = 0.5 \, M \times 0.035 \, L = 0.0175 \, moles \]

### Step 2: Determine the number of moles of \( K_2Cr_2O_7 \) required

The oxidation reaction involves the conversion of \( Fe^{2+} \) to \( Fe^{3+} \). The balanced equation shows that 1 mole of \( K_2Cr_2O_7 \) (which contains 6 moles of \( Cr_2O_7^{2-} \)) can oxidize 6 moles of \( Fe^{2+} \).

Thus, the stoichiometry is:

\[ 1 \, \text{mol of } K_2Cr_2O_7 \text{ oxidizes } 6 \, \text{mol of } FeSO_4 \]

From this, we can find the moles of \( K_2Cr_2O_7 \) needed:

\[ \text{Moles of } K_2Cr_2O_7 = \frac{\text{Moles of } FeSO_4}{6} \]

\[ \text{Moles of } K_2Cr_2O_7 = \frac{0.0175}{6} \approx 0.00291667 \, moles \]

### Step 3: Calculate the volume of \( K_2Cr_2O_7 \) solution required

Now, we can calculate the volume of the 0.1 M \( K_2Cr_2O_7 \) solution needed to provide the required moles.

Using the formula:

\[ \text{Volume (L)} = \frac{\text{Moles}}{\text{Molarity}} \]

Given:
- Molarity of \( K_2Cr_2O_7 = 0.1 \, M \)

\[ \text{Volume of } K_2Cr_2O_7 = \frac{0.00291667}{0.1} = 0.0291667 \, L \]

Converting to mL:

\[ \text{Volume of } K_2Cr_2O_7 = 0.0291667 \times 1000 \approx 29.17 \, mL \]

### Final Answer

The volume of 0.1 M \( K_2Cr_2O_7 \) required to oxidize 35 mL of 0.5 M \( FeSO_4 \) solution is approximately **29.17 mL**.

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