Calculate the amount of sodium chloride (electrolyte) which must be added to one kilogram of water so that the freezing point is depressed by 3K. Give `k_(f)` for water = `1.86 K kg mol^(-1)`

Explanation

To solve the problem of calculating the amount of sodium chloride (NaCl) that must be added to one kilogram of water to depress the freezing point by 3K, we will follow these steps:

### Step 1: Understand the formula for freezing point depression
The formula for freezing point depression is given by:
\[ \Delta T_f = K_f \times m \times i \]
Where:
- \(\Delta T_f\) = depression in freezing point (in Kelvin)
- \(K_f\) = cryoscopic constant (for water, \(K_f = 1.86 \, \text{K kg mol}^{-1}\))
- \(m\) = molality of the solution (in mol/kg)
- \(i\) = van 't Hoff factor (for NaCl, \(i = 2\) since it dissociates into Na\(^+\) and Cl\(^-\))

### Step 2: Substitute the known values into the formula
Given:
- \(\Delta T_f = 3 \, \text{K}\)
- \(K_f = 1.86 \, \text{K kg mol}^{-1}\)
- \(i = 2\)

Substituting these values into the formula:
\[ 3 = 1.86 \times m \times 2 \]

### Step 3: Solve for molality (m)
Rearranging the equation to solve for \(m\):
\[ m = \frac{3}{1.86 \times 2} \]
Calculating:
\[ m = \frac{3}{3.72} \approx 0.806 \, \text{mol/kg} \]

### Step 4: Calculate the number of moles of NaCl needed
Molality is defined as:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]
Given that the mass of the solvent (water) is 1 kg, we can express the moles of NaCl as:
\[ \text{moles of NaCl} = m \times \text{mass of solvent} = 0.806 \, \text{mol/kg} \times 1 \, \text{kg} = 0.806 \, \text{mol} \]

### Step 5: Convert moles of NaCl to grams
To find the mass of NaCl needed, we use the molar mass of NaCl:
- Molar mass of NaCl = 58.5 g/mol

Calculating the mass:
\[ \text{mass of NaCl} = \text{moles of NaCl} \times \text{molar mass of NaCl} = 0.806 \, \text{mol} \times 58.5 \, \text{g/mol} \approx 47.15 \, \text{g} \]

### Final Answer
The amount of sodium chloride (NaCl) that must be added to one kilogram of water to depress the freezing point by 3K is approximately **47.15 grams**.

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