0.1 M KI and 0.2 M `AgNO_(3)` are mixed in 3 : 1 volume ratio. The depression of freezing point of the resulting solution will be [`K_(b)(H_(2)O) = 1.86 K kg "mol"^(-1)`]:

Options

A. 3.72 K
B. 1.86 K
C. 0.93 K
D. 0.279 K

Answer

To solve the problem of finding the depression of freezing point when mixing 0.1 M KI and 0.2 M AgNO3 in a 3:1 volume ratio, we can follow these steps:

### Step 1: Determine the moles of KI and AgNO3
Given:
- Concentration of KI = 0.1 M
- Concentration of AgNO3 = 0.2 M
- Volume ratio = 3:1

Assuming we take 3 L of KI and 1 L of AgNO3 for simplicity:
- Moles of KI = 0.1 M × 3 L = 0.3 moles
- Moles of AgNO3 = 0.2 M × 1 L = 0.2 moles

### Step 2: Determine the reaction
When KI reacts with AgNO3, the reaction is:
\[ \text{KI} + \text{AgNO}_3 \rightarrow \text{AgI} \downarrow + \text{KNO}_3 \]

From the reaction:
- 0.2 moles of AgNO3 will react with 0.2 moles of KI, producing 0.2 moles of AgI and leaving:
- Remaining KI = 0.3 moles - 0.2 moles = 0.1 moles

### Step 3: Calculate the total moles of particles in solution
After the reaction, we have:
- Remaining KI = 0.1 moles
- KNO3 produced = 0.2 moles (from the reaction)
- Total moles of ions:
- From KI: 0.1 moles of KI produces 0.1 moles of K⁺ and 0.1 moles of I⁻
- From KNO3: 0.2 moles of KNO3 produces 0.2 moles of K⁺ and 0.2 moles of NO3⁻

Total moles of ions:
- K⁺ = 0.1 + 0.2 = 0.3 moles
- I⁻ = 0.1 moles
- NO3⁻ = 0.2 moles
- Total = 0.3 + 0.1 + 0.2 = 0.6 moles of particles

### Step 4: Calculate the total volume of the solution
Total volume = 3 L (KI) + 1 L (AgNO3) = 4 L

### Step 5: Calculate molality (m)
Molality (m) is defined as moles of solute per kg of solvent. In this case, we will use the total volume to find the concentration:
\[ m = \frac{\text{Total moles of particles}}{\text{Total volume in L}} = \frac{0.6 \text{ moles}}{4 \text{ L}} = 0.15 \text{ moles/L} \]

### Step 6: Calculate the depression of freezing point (ΔTf)
Using the formula:
\[ \Delta T_f = K_f \cdot m \]
Where:
- \( K_f \) for water = 1.86 K kg/mol
- m = 0.15 mol/kg (since we are considering the moles in the total volume)

Calculating:
\[ \Delta T_f = 1.86 \cdot 0.15 = 0.279 \text{ K} \]

### Final Answer
The depression of the freezing point of the resulting solution is **0.279 K**.

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