1. Describe the enthalpy changes during the dissolving of KBr into C 2 H 6 : The

Question

1. Describe the enthalpy changes during the dissolving of KBr into C2H6:

The enthalpy of separating the solute is slightly endothermic; the enthalpy of separating the solvent is slightly endothermic; the enthalpy of mixing is highly exothermic; the enthalpy of solution is highly exothermic.

2. What will be the boiling point of a 5 molal solution of LiF? Kb for water = 0.512 C/m.

A. The enthalpy of separating the solute is highly endothermic; the enthalpy of separating the solvent is slightly endothermic; the enthalpy of mixing is slightly exothermic; the enthalpy of solution is highly endothermic. B. The enthalpy of separating the solute is slightly endothermic; the enthalpy of separating the solvent is highly endothermic; the enthalpy of mixing is slightly exothermic; the enthalpy of solution is highly endothermic. C. The enthalpy of separating the solute is highly endothermic; the enthalpy of separating the solvent is slightly endothermic; the enthalpy of mixing is highly exothermic; the enthalpy of solution is slightly endothermic. D.

The enthalpy of separating the solute is slightly endothermic; the enthalpy of separating the solvent is slightly endothermic; the enthalpy of mixing is highly exothermic; the enthalpy of solution is highly exothermic.

Explanation / Answer

1. The answer is:

(A) The enthalpy of separating the solute is highly endothermic; the enthalpy of separating the solvent is slightly endothermic; the enthalpy of mixing is slightly exothermic; the enthalpy of solution is highly endothermic.

Separating ionic KBr solute => high energy needed => highly endothermic

Separating molecular C2H6 solvent (dispersion forces) => low energy needed => slightly endothermic

Mixing incompatible bond types (ionic and dispersion forces) => low energy released => slightly exothermic

Enthalpy of solution = sum of energies above = highly endothermic

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2. The answer is: A. 105.1 deg C

LiF => Li+ + F-

van't Hoff factor i = 2

Boiling point elevation DTb = i x Kb x molality

= 2 x 0.512 x 5 = 5.12 deg C

Boiling point of solution = boiling point of water + DTb

= 100 + 5.12 = 105.12 deg C = 105.1 deg C

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