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)`]:
(a)3.72 K
(b)1.86 K
(c)0.93 K
(d)0.279 K

Options

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

Answer

To solve the problem of calculating the depression of the 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 liters of KI and 1 liter of AgNO3:

- Moles of KI = Concentration × Volume = 0.1 mol/L × 3 L = 0.3 moles
- Moles of AgNO3 = Concentration × Volume = 0.2 mol/L × 1 L = 0.2 moles

### Step 2: Write the chemical reaction
When KI reacts with AgNO3, the following reaction occurs:
\[ \text{KI} + \text{AgNO}_3 \rightarrow \text{AgI (s)} + \text{KNO}_3 \]

From the reaction:
- 0.2 moles of AgNO3 will react with 0.2 moles of KI to produce 0.2 moles of AgI (which precipitates out) and 0.2 moles of KNO3.

### Step 3: Calculate the remaining moles after the reaction
After the reaction:
- Remaining moles of KI = 0.3 moles (initial) - 0.2 moles (reacted) = 0.1 moles
- Moles of KNO3 produced = 0.2 moles

### Step 4: Calculate the total moles of ions in solution
The ions present in the solution are:
- From KNO3: 0.2 moles of KNO3 dissociates into 0.2 moles of K⁺ and 0.2 moles of NO3⁻ (total = 0.4 moles)
- From remaining KI: 0.1 moles of KI dissociates into 0.1 moles of K⁺ and 0.1 moles of I⁻ (total = 0.2 moles)

Total moles of ions:
- K⁺: 0.2 (from KNO3) + 0.1 (from KI) = 0.3 moles
- NO3⁻: 0.2 moles
- I⁻: 0.1 moles

Total moles of ions = 0.3 + 0.2 + 0.1 = 0.6 moles

### Step 5: Calculate the total volume of the solution
Total volume = 3 L (KI) + 1 L (AgNO3) = 4 L

### Step 6: Calculate the molality of the solution
Molality (m) is calculated as:
\[ m = \frac{\text{Total moles of solute}}{\text{Volume of solvent in kg}} \]

Since we are considering the total volume of the solution (4 L), we can assume the density of the solution is approximately 1 kg/L for simplicity, hence:
- Volume of solvent = 4 kg (approx.)

So, molality:
\[ m = \frac{0.6 \text{ moles}}{4 \text{ kg}} = 0.15 \text{ mol/kg} \]

### Step 7: Calculate the depression of the freezing point
Using the formula for depression of freezing point:
\[ \Delta T_f = K_f \cdot m \]
Where \( K_f \) for water is given as 1.86 K kg/mol.

Substituting the values:
\[ \Delta T_f = 1.86 \, \text{K kg/mol} \times 0.15 \, \text{mol/kg} = 0.279 \, \text{K} \]

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

### Final Answer
(d) 0.279 K

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