A solution that contains 2.00 g of ferrocene [molar mass = 186.0 g/mol] dissolved in 50.0 g of ethylene bromide (C₂H₄Br₂) has a freezing point of 7.26 ° C. The normal freezing point of ethylene bromide is 9.80 ° C. What is the freezing point depression constant for this solvent?

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The freezing point depression ($\Delta T_f$) is given by the equation $\Delta T_f = i \cdot K_f \cdot m$, where $i$ is the van't Hoff factor (which is 1 for non-electrolytes like ferrocene), $K_f$ is the freezing point depression constant, and $m$ is the molality of the solution. Given:- Mass of ferrocene = 2.00 g- Molar mass of ferrocene = 186.0 g/mol- Mass of ethylene bromide = 50.0 g (solvent)- Freezing point of solution = 7.26 °C- Normal freezing point of ethylene bromide = 9.80 °CFirst, calculate the molality ($m$) of the solution:$$ \text{Moles of ferrocene} = \frac{\text{Mass of ferrocene}}{\text{Molar mass of ferrocene}} = \frac{2.00 \, \text{g}}{186.0 \, \text{g/mol}} = 0.01075 \, \text{mol} $$$$ \text{Mass of solvent in kg} = 50.0 \, \text{g} = 0.0500 \, \text{kg} $$$$ \text{Molality} (m) = \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}} = \frac{0.01075 \, \text{mol}}{0.0500 \, \text{kg}} = 0.215 \, \text{mol/kg} $$Next, calculate the freezing point depression ($\Delta T_f$):$$ \Delta T_f = \text{Normal freezing point} - \text{Freezing point of solution} = 9.80 \, °C - 7.26 \, °C = 2.54 \, °C $$Finally, solve for $K_f$:$$ K_f = \frac{\Delta T_f}{i \cdot m} = \frac{2.54 \, °C}{1 \cdot 0.215 \, \text{mol/kg}} = 11.8 \, °C/m $$Therefore, the freezing point depression constant for this solvent is 11.8 °C/m.

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