Calculate the amount of `KCl` which must be added to `1kg` of water so that the freezing point is depressed by `2K`. (`K_(f)` for water `=1.86K kg "mol"^(-1)`).

Answer

To solve the problem of calculating the amount of KCl that must be added to 1 kg of water to depress the freezing point by 2 K, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Freezing point depression (ΔTf) = 2 K
- Cryoscopic constant (Kf) for water = 1.86 K kg/mol
- Mass of solvent (water) = 1 kg

2. **Use the Freezing Point Depression Formula:**
The formula for freezing point depression is given by:
\[ \Delta T_f = i \cdot K_f \cdot m \]
where:
- \( \Delta T_f \) = freezing point depression
- \( i \) = van 't Hoff factor (number of particles the solute breaks into)
- \( K_f \) = cryoscopic constant
- \( m \) = molality of the solution (moles of solute per kg of solvent)

3. **Determine the Van 't Hoff Factor (i) for KCl:**
KCl dissociates into two ions in solution:
\[ KCl \rightarrow K^+ + Cl^- \]
Therefore, the van 't Hoff factor \( i \) for KCl is 2.

4. **Rearrange the Freezing Point Depression Formula:**
Substitute the known values into the formula:
\[ 2 = 2 \cdot 1.86 \cdot m \]
Now, solve for \( m \):
\[ m = \frac{2}{2 \cdot 1.86} = \frac{2}{3.72} \approx 0.5376 \text{ mol/kg} \]

5. **Calculate the Number of Moles of KCl:**
Since we have 1 kg of water, the number of moles of KCl needed is equal to the molality:
\[ \text{Moles of KCl} = m \cdot \text{mass of solvent} = 0.5376 \text{ mol/kg} \cdot 1 \text{ kg} = 0.5376 \text{ moles} \]

6. **Calculate the Molar Mass of KCl:**
The molar mass of KCl can be calculated as follows:
- Atomic mass of K = 39 g/mol
- Atomic mass of Cl = 35.5 g/mol
\[ \text{Molar mass of KCl} = 39 + 35.5 = 74.5 \text{ g/mol} \]

7. **Calculate the Mass of KCl Required:**
Now, calculate the mass of KCl using the number of moles and the molar mass:
\[ \text{Mass of KCl} = \text{Moles of KCl} \cdot \text{Molar mass of KCl} = 0.5376 \text{ moles} \cdot 74.5 \text{ g/mol} \approx 40.05 \text{ g} \]

### Final Answer:
The amount of KCl that must be added to 1 kg of water to depress the freezing point by 2 K is approximately **40.05 g**.

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