How many millimoles of potassium dichromate is required to oxidise 24 ml of 0.5 M Mohr salt solution in acidic medium ?

Explanation

To determine how many millimoles of potassium dichromate (K2Cr2O7) are required to oxidize 24 mL of 0.5 M Mohr salt (ferrous ammonium sulfate, (NH4)2Fe(SO4)2·6H2O) solution in acidic medium, we can follow these steps:

### Step 1: Calculate the number of moles of Mohr salt
First, we need to calculate the number of moles of Mohr salt in the given volume and concentration.

\[ \text{Moles of Mohr salt} = \text{Concentration (M)} \times \text{Volume (L)} \]

Given:
- Concentration = 0.5 M
- Volume = 24 mL = 0.024 L

\[ \text{Moles of Mohr salt} = 0.5 \, \text{mol/L} \times 0.024 \, \text{L} = 0.012 \, \text{mol} \]

### Step 2: Determine the stoichiometry of the reaction
In acidic medium, the reaction between potassium dichromate and Mohr salt can be represented as follows:

\[ K_2Cr_2O_7 + 6Fe^{2+} + 14H^+ \rightarrow 2Cr^{3+} + 6Fe^{3+} + 7H_2O \]

From the balanced equation, we see that 1 mole of potassium dichromate (K2Cr2O7) reacts with 6 moles of ferrous ions (Fe²⁺).

### Step 3: Calculate the moles of potassium dichromate required
Using the stoichiometry from the balanced equation, we can find out how many moles of potassium dichromate are needed for the moles of Mohr salt we calculated.

\[ \text{Moles of K}_2Cr_2O_7 = \frac{\text{Moles of Fe}^{2+}}{6} \]

Since each mole of Mohr salt provides 1 mole of Fe²⁺, we have:

\[ \text{Moles of K}_2Cr_2O_7 = \frac{0.012 \, \text{mol}}{6} = 0.002 \, \text{mol} \]

### Step 4: Convert moles of potassium dichromate to millimoles
To convert moles to millimoles, we multiply by 1000:

\[ \text{Millimoles of K}_2Cr_2O_7 = 0.002 \, \text{mol} \times 1000 = 2 \, \text{mmol} \]

### Final Answer
**2 millimoles of potassium dichromate are required to oxidize 24 mL of 0.5 M Mohr salt solution in acidic medium.**

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