**Question 3: Freezing Point Depression Calculation****Problem Statement:**What is the freezing point of a 3.0 m (molal) aqueous solution of NaCl (Sodium Chloride)? The freezing point depression constant ($K_f$) for water is 1.86°C kg solvent/mol solute.**Options:**a) 5.6°C b) -5.6°C c) 11°C d) -11°C### Explanation:To solve this problem, we'll use the freezing point depression formula:\[ \Delta T_f = i \times K_f \times m \]where:- $\Delta T_f$ is the freezing point depression,- $i$ is the van't Hoff factor (which represents the number of particles the solute splits into),- $K_f$ is the freezing point depression constant,For NaCl, which dissociates into two ions (Na$^+$ and Cl$^-$), the van't Hoff factor ($i$) is 2.Given:- $K_f = 1.86^\circ$C kg/mol,- $m = 3.0$ mol/kg (molality).Plug these values into the formula to find $\Delta T_f$:\[ \Delta T_f = 2 \times 1.86 \,^\circ\text{C kg/mol} \times 3.0 \,\text{mol/kg} \]\[ \Delta T_f = 2 \times 5.58 \,^\circ\text{C} \]\[ \Delta T_f = 11.16 \,^\circ\text{C} \]The freezing point of pure water is 0°C. The freezing point of the solution will be lowered by 11.16°C from pure water's freezing point. Thus, the new freezing point is:\[ 0^\circ\text{C} - 11.16^\circ\text{C} = -11.16^\circ\text{C} \]However, the closest match to this calculation in the options provided is:**d) -11°C**

Freezing point depression is the phenomena that describes why adding a solute to a solvent results…

Answered: 3) What is the freezing point of a 3.0…

3) What is the freezing point of a 3.0 m aqueous solution of NaCl? (The freezing point depression constant for water is 1.86 °C kg solvent/mol solute.) a) 5.6 °C b) -5.6 °C c) 11°C d) -11°C

Chemistry

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ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

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