To solve the problem of how much NaCl should be added to 500 g of water to decrease the freezing point to -0.3°C, we will use the formula for freezing point depression:
\[
\Delta T_f = i \cdot K_f \cdot m
\]
Where:
- \(\Delta T_f\) = depression in freezing point
- \(i\) = van 't Hoff factor (number of particles the solute dissociates into)
- \(K_f\) = freezing point depression constant
- \(m\) = molality of the solution
### Step 1: Calculate the depression in freezing point (\(\Delta T_f\))
The freezing point of pure water is 0°C. The desired freezing point of the solution is -0.3°C.
\[
\Delta T_f = T_f^{\text{solvent}} - T_f^{\text{solution}} = 0°C - (-0.3°C) = 0.3°C
\]
### Step 2: Identify the van 't Hoff factor (\(i\))
For NaCl, which dissociates into Na⁺ and Cl⁻ ions, the van 't Hoff factor \(i\) is:
\[
i = 2
\]
### Step 3: Use the freezing point depression constant (\(K_f\))
The freezing point depression constant for water is given as:
\[
K_f = 2 \, \text{K kg mol}^{-1}
\]
### Step 4: Set up the equation for molality (\(m\))
Rearranging the freezing point depression formula gives us:
\[
m = \frac{\Delta T_f}{i \cdot K_f}
\]
Substituting the known values:
\[
m = \frac{0.3}{2 \cdot 2} = \frac{0.3}{4} = 0.075 \, \text{mol/kg}
\]
### Step 5: Calculate the mass of the solvent in kg
The mass of the solvent (water) is given as 500 g. We need to convert this to kg:
\[
\text{mass of solvent} = 500 \, \text{g} = 0.5 \, \text{kg}
\]
### Step 6: Calculate the number of moles of NaCl needed
Using the definition of molality:
\[
m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}
\]
Rearranging gives:
\[
\text{moles of solute} = m \cdot \text{mass of solvent in kg} = 0.075 \, \text{mol/kg} \cdot 0.5 \, \text{kg} = 0.0375 \, \text{mol}
\]
### Step 7: Calculate the mass of NaCl needed
The molar mass of NaCl is:
\[
\text{Molar mass of NaCl} = 23 \, \text{g/mol (Na)} + 35.5 \, \text{g/mol (Cl)} = 58.5 \, \text{g/mol}
\]
Now, calculate the mass of NaCl required:
\[
\text{mass of NaCl} = \text{moles of NaCl} \cdot \text{molar mass of NaCl} = 0.0375 \, \text{mol} \cdot 58.5 \, \text{g/mol} = 2.19375 \, \text{g}
\]
### Step 8: Round the answer
Rounding to two significant figures, we find:
\[
\text{mass of NaCl} \approx 2.19 \, \text{g}
\]
### Final Answer
The amount of NaCl that should be added to 500 g of water to decrease the freezing point to -0.3°C is approximately **2.19 g**.
---
