Which lowers the freezing point of 2.0 kg of water more, 0.20 mol NaOH or 0.20 mol Ba(OH)2 ?

step 1 In this problem we are given B A .O .H produces three ions of solution as you can see and N .A .

Which lowers the freezing point of 2.0 kg of water more,...

Which lowers the freezing point of $2.0 \mathrm{~kg}$ of water more, $0.20 \mathrm{~mol} \mathrm{NaOH}$ or $0.20 \mathrm{~mol} \mathrm{Ba}(\mathrm{OH})_{2} ?$ Both compounds are strong electrolytes. Explain.

Key Concepts

Electrolyte Dissociation

Electrolytes completely dissociate into ions in solution, which directly influences the number of particles available. Strong electrolytes produce more ions per formula unit compared to non-electrolytes, thereby enhancing the colligative effects such as freezing point depression, since the effect is contingent on the total number of dissolved particles.

Colligative Properties

These properties depend on the number of solute particles dissolved in a solvent, rather than on the chemical nature of the solute. Changes such as freezing point depression and boiling point elevation are affected by factors like the concentration and dissociation of the solute particles in the solution.

Freezing Point Depression

This concept describes the lowering of a solvent's freezing point when a solute is dissolved in it. The decrease in freezing point is proportional to the solution’s molal concentration of dissolved particles, making it a key technique for studying how solutes (especially electrolytes) alter phase transitions in a solvent.

van't Hoff Factor

The van't Hoff factor (i) represents the number of particles a solute splits into upon dissolution. In the context of freezing point depression and other colligative properties, a higher van't Hoff factor indicates more particles in the solution, leading to a larger effect on physical properties such as the freezing point.

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