`30 mL` of `0.1M KI(aq)` and `10 mL` of `0.2M AgNO_(3)` are mixed. The solution is then filtered out. Assuming that no change in total volume, the resulting solution will freezing at:
`[K_(f) "for" H_(2)O = 1.86K kg mol^(-1)`, assume molality = molarity]

Answer

To solve the problem step by step, we will follow the process of mixing the solutions, determining the amount of solute remaining, and calculating the freezing point depression.

### Step 1: Calculate the moles of KI and AgNO₃
- **Moles of KI**:
\[ \text{Molarity} = \frac{\text{moles}}{\text{volume in L}} \implies \text{moles} = \text{Molarity} \times \text{volume in L} \]
\[ \text{Moles of KI} = 0.1 \, \text{mol/L} \times 0.030 \, \text{L} = 0.003 \, \text{mol} = 3 \, \text{mmol} \]

- **Moles of AgNO₃**:
\[ \text{Moles of AgNO₃} = 0.2 \, \text{mol/L} \times 0.010 \, \text{L} = 0.002 \, \text{mol} = 2 \, \text{mmol} \]

### Step 2: Determine the reaction and remaining moles
When KI reacts with AgNO₃, they form AgI (which precipitates) and KNO₃. The balanced reaction is:
\[ \text{KI} + \text{AgNO}_3 \rightarrow \text{AgI} \downarrow + \text{KNO}_3 \]

- **Reactants**:
- 3 mmol of KI
- 2 mmol of AgNO₃

Since 1 mole of KI reacts with 1 mole of AgNO₃, 2 mmol of KI will react with 2 mmol of AgNO₃, leaving:
- Remaining KI = 3 mmol - 2 mmol = 1 mmol
- All AgNO₃ is consumed.

### Step 3: Calculate the total volume of the solution
The total volume after mixing is:
\[ 30 \, \text{mL} + 10 \, \text{mL} = 40 \, \text{mL} \]

### Step 4: Calculate the total moles of solute remaining in solution
Remaining solutes in solution:
- 1 mmol of KI
- 2 mmol of KNO₃ (produced from the reaction)

Total moles of solute in solution:
\[ 1 \, \text{mmol (KI)} + 2 \, \text{mmol (KNO}_3\text{)} = 3 \, \text{mmol} \]

### Step 5: Calculate the molality of the solution
Assuming the density of the solution is similar to water, the mass of the solvent (water) can be approximated:
\[ \text{Mass of water} = \text{Volume} \times \text{Density} \approx 40 \, \text{mL} \times 1 \, \text{g/mL} = 40 \, \text{g} = 0.040 \, \text{kg} \]

Now, calculate molality (moles of solute per kg of solvent):
\[ \text{Molality} = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.003 \, \text{mol}}{0.040 \, \text{kg}} = 0.075 \, \text{mol/kg} \]

### Step 6: Calculate the freezing point depression
Using the formula:
\[ \Delta T_f = i \cdot K_f \cdot m \]
Where:
- \(i\) (van 't Hoff factor) for KI = 2 (K⁺ and I⁻)
- \(i\) for KNO₃ = 2 (K⁺ and NO₃⁻)
- Total \(i = 2 + 2 = 4\)

Substituting the values:
\[ \Delta T_f = 4 \cdot 1.86 \, \text{K kg mol}^{-1} \cdot 0.075 \, \text{mol/kg} = 0.558 \, \text{K} \]

### Step 7: Calculate the freezing point of the solution
The freezing point of pure water is 0 °C. Therefore:
\[ \text{Freezing point of solution} = 0 - 0.558 = -0.558 \, \text{°C} \]

### Final Answer
The resulting solution will freeze at approximately **-0.558 °C**.

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