25 ml of `FeC_2O_4` dissolved in 186 gm of water calculate depression in freezing point. It 10 ml of same `FeC_2O_4` titrated with 30 ml of 0.4 M `KMnO_4` in acidic medium (`k_f` for `H_2O=1.86`,Assume 100% ionisation of `FeC_2O_4`).

Explanation

To solve the problem, we need to calculate the depression in freezing point (ΔTf) of a solution of iron(II) oxalate (FeC₂O₄) dissolved in water. We will follow these steps:

### Step 1: Determine the number of moles of FeC₂O₄ in the titration
We know that 10 mL of FeC₂O₄ is titrated with 30 mL of 0.4 M KMnO₄ in acidic medium.

1. Calculate the equivalents of KMnO₄:
\[ \text{Equivalents of KMnO}_4 = \text{Volume (L)} \times \text{Molarity} \times \text{n (number of electrons transferred)} \]
For KMnO₄ in acidic medium, n = 5 (it gains 5 electrons).
\[ \text{Volume} = 30 \text{ mL} = 0.030 \text{ L} \]
\[ \text{Equivalents of KMnO}_4 = 0.030 \times 0.4 \times 5 = 0.060 \text{ equivalents} \]

2. Write the relationship between the equivalents of FeC₂O₄ and KMnO₄:
\[ \text{Equivalents of FeC}_2\text{O}_4 = \text{Equivalents of Fe}^{2+} + \text{Equivalents of C}_2\text{O}_4^{2-} \]
Since each mole of FeC₂O₄ produces 1 mole of Fe²⁺ and 1 mole of C₂O₄²⁻, we have:
\[ \text{Equivalents of FeC}_2\text{O}_4 = n + 2n = 3n \]
Setting this equal to the equivalents of KMnO₄:
\[ 3n = 0.060 \implies n = \frac{0.060}{3} = 0.020 \text{ moles} \]

### Step 2: Calculate the molarity of FeC₂O₄
Using the volume of FeC₂O₄ solution used in the titration (10 mL):
\[ \text{Molarity (M)} = \frac{n}{\text{Volume (L)}} = \frac{0.020}{0.010} = 2 \text{ M} \]

### Step 3: Calculate the total moles of FeC₂O₄ in 25 mL solution
Using the molarity calculated:
\[ \text{Total moles in 25 mL} = \text{Molarity} \times \text{Volume (L)} = 2 \times 0.025 = 0.050 \text{ moles} \]

### Step 4: Calculate the depression in freezing point (ΔTf)
Using the formula:
\[ \Delta T_f = i \cdot K_f \cdot m \]
Where:
- \( i \) = van 't Hoff factor (number of particles in solution) = 3 (1 Fe²⁺ and 1 C₂O₄²⁻)
- \( K_f \) for water = 1.86 °C kg/mol
- \( m \) = molality = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}}

Convert the mass of water from grams to kilograms:
\[ 186 \text{ g} = 0.186 \text{ kg} \]
Calculate molality:
\[ m = \frac{0.050 \text{ moles}}{0.186 \text{ kg}} \approx 0.269 \text{ mol/kg} \]

Now substitute into the ΔTf formula:
\[ \Delta T_f = 3 \cdot 1.86 \cdot 0.269 \approx 1.50 \text{ °C} \]

### Final Answer
The depression in freezing point is approximately **1.50 °C**.

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