Suppose 115 g of KCl is dissolved in 750 mL of water (density = 1.005 g/mL). What would be the...
Question:
Suppose 115 g of KCl is dissolved in 750 mL of water (density = 1.005 g/mL). What would be the freezing point and boiling point of that solution, assuming the {eq}K_f {/eq} of water is 1.86 {eq}^{\circ} {/eq}C/m and {eq}K_b {/eq} is 0.512 {eq}^{\circ} {/eq}C/m? (Assume that KCl fully dissociates with no pairing of ions.)
Solution's Boiling and Freezing Points:
The presence of the solute affects the boiling point and freezing point of the solution. The freezing point of a solution is lower than the freezing point of the pure solvent. The difference between these freezing points gives the freezing-point depression of the solution. On the other hand, the boiling point of a solution is higher than the boiling point of the pure solvent. The difference between these boiling points determines the boiling-point elevation of the solution.
Answer and Explanation: 1
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View this answerThe solution in this problem is made up of the solute potassium chloride, KCl, and solvent water. To determine the freezing point and boiling point of...
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Colligative Properties & Raoult's Law | Equations & Examples
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Chapter 8 / Lesson 8
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Learn about colligative properties and their equations. See Raoult's law in vapor pressure, a colligative property, and osmotic pressure problem examples. Learn about freezing point depression and boiling point elevation.
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