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Step-by-Step Solution
Step 1: Identify the Relevant Formula
The depression in freezing point (ΔTf) for an ionic solute is given by the equation:
\Delta T_f = i \, k_f \, m
\Delta T_f = Depression in freezing point
i = Van’t Hoff factor
k_f = Molal freezing point depression constant
m = Molality of the solution
Step 2: Note the Given Data
Desired depression in freezing point, \Delta T_f = 0.2^\circ \text{C}
Mass of water, m_{\text{water}} = 600 \,\text{g}
k_f (for water) = 2\,\text{K}\,\text{kg}\,\text{mol}^{-1}
Van’t Hoff factor for NaCl, i = 2
Molar mass of NaCl = 58.5\,\text{g/mol}
Step 3: Express Molality
Molality (m) is defined as the number of moles of solute per kilogram of solvent. Here, the mass of solvent (water) is 600\,\text{g} = 0.600\,\text{kg} . If W_{\text{NaCl}} grams of NaCl are required, the number of moles of NaCl is \frac{W_{\text{NaCl}}}{58.5} . Hence,
m = \frac{\frac{W_{\text{NaCl}}}{58.5}}{0.600} \,\text{mol/kg} = \frac{W_{\text{NaCl}} \times 1000}{58.5 \times 600} \,\text{mol/kg}.
Step 4: Apply the Freezing Point Depression Formula
Substitute the known values into \Delta T_f = i \, k_f \, m :
0.2 = 2 \times 2 \times \frac{W_{\text{NaCl}} \times 1000}{58.5 \times 600}.
Here, the factor 2 (for i) multiplies the other 2 (for k_f =2 and the expression for m).
Step 5: Solve for W_{\text{NaCl}}
Rearrange and solve for W_{\text{NaCl}} :
0.2 = 4 \times \frac{W_{\text{NaCl}} \times 1000}{58.5 \times 600} \\[6pt]
\Rightarrow W_{\text{NaCl}} = \frac{0.2 \times 58.5 \times 600}{4 \times 1000}.
Performing the calculation:
W_{\text{NaCl}} = \frac{0.2 \times 58.5 \times 600}{4000} = 1.76\,\text{g}.
Step 6: Final Answer
The amount of NaCl that must be added to 600 g of water to lower its freezing point by 0.2 °C is 1.76 g.
