0.84 g iron ore containing x percent of iron was taken in a solution containing all the iron in ferrous condition. The solution required x ml of a dichromatic solution for oxidizing the iron content to ferric state. Calculate the strength of dichromatic solution.

Answer

To solve the problem step by step, we will follow the information provided and apply the concepts of stoichiometry and normality in redox reactions.

### Step 1: Determine the mass of iron in the iron ore
Given that the iron ore weighs 0.84 g and contains x percent of iron, we can calculate the mass of iron in the ore.

\[ \text{Mass of iron} = \frac{x}{100} \times 0.84 \text{ g} \]

### Step 2: Calculate the moles of ferrous iron (Fe²⁺)
The molar mass of iron (Fe) is 56 g/mol. To find the moles of ferrous iron, we use the formula:

\[ \text{Moles of Fe}^{2+} = \frac{\text{Mass of iron}}{\text{Molar mass of Fe}} = \frac{\frac{x}{100} \times 0.84}{56} \]

### Step 3: Write the half-reaction for the oxidation of ferrous iron
The oxidation of ferrous iron (Fe²⁺) to ferric iron (Fe³⁺) can be represented as:

\[ \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^- \]

This indicates that 1 mole of Fe²⁺ loses 1 mole of electrons.

### Step 4: Determine the stoichiometry with dichromate (Cr₂O₇²⁻)
The dichromate ion (Cr₂O₇²⁻) is reduced in the following reaction:

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

From this reaction, we can see that 1 mole of Cr₂O₇²⁻ reacts with 6 moles of electrons.

### Step 5: Relate moles of Fe²⁺ to moles of Cr₂O₇²⁻
Let the moles of Fe²⁺ be \( n \). Since 1 mole of Fe²⁺ corresponds to 1 mole of electrons, the moles of Cr₂O₇²⁻ required will be:

\[ \text{Moles of Cr}_2\text{O}_7^{2-} = \frac{n}{6} \]

### Step 6: Calculate the volume of dichromate solution used
Given that the solution required x ml of dichromate solution, we can express the normality (N) of the dichromate solution as:

\[ N = \frac{\text{Moles of solute}}{\text{Volume of solution in liters}} \]

Substituting the moles of Cr₂O₇²⁻ we found earlier:

\[ N = \frac{\frac{n}{6}}{\frac{x}{1000}} = \frac{n \times 1000}{6x} \]

### Step 7: Substitute for n (moles of Fe²⁺)
Now, substituting \( n \) from our earlier calculation:

\[ n = \frac{\frac{x}{100} \times 0.84}{56} \]

Substituting this back into the normality equation gives:

\[ N = \frac{\left(\frac{\frac{x}{100} \times 0.84}{56}\right) \times 1000}{6x} \]

### Step 8: Simplify the expression
Now we simplify the expression to find the normality:

\[ N = \frac{0.84 \times 1000}{56 \times 6 \times 100} \]

Calculating this gives:

\[ N = \frac{840}{33600} = 0.025 \text{ N} \]

### Step 9: Calculate the strength of the dichromate solution
The strength of the dichromate solution can be given in terms of normality:

\[ \text{Strength} = N \times \text{Equivalent weight of K}_2\text{Cr}_2\text{O}_7 \]

The equivalent weight of K₂Cr₂O₇ is 49 (molar mass 294 g/mol divided by n factor 6). Thus:

\[ \text{Strength} = 0.025 \times 49 = 1.225 \text{ g/L} \]

### Final Answer
The strength of the dichromate solution is **1.225 g/L**.

---