A $\pu{1.50 g}$ sample of $\ce{KCl}$ is added to $\pu{35.0 g}$ $\ce{H2O}$ in a styrofoam cup and stirred until dissolved. The temperature of the solution drops from $24.8$ to $\pu{22.4 ^\circ C}$. Assume that the specific heat and density of the resulting solution are equal to those of water, $\pu{4.18 J g-1 ^\circ C-1}$ and $\pu{1.00 g mL-1}$, respectively, and assume that no heat is lost to the calorimeter itself, nor to the surroundings.

$$\ce{KCl (s) + H2O (l) -> KCl (aq)} \qquad \Delta H = ?$$

a) (2 points) Is the reaction endothermic or exothermic (circle the correct answer)?


b) (4 points) What is the heat of solution of $\ce{KCl}$ expressed in kilojoules per mole of $\ce{KCl}$?

$$q_\mathrm{rxn} = -q_\mathrm{cal}$$

I multiplied the sample $\pu{1.50 g}$ by $\pu{4.18 J} \cdot (-2.4) = \pu{-15.048 J}$

Divided that by $1000 = -0.015048$; thus, $0.015048$

However, my answer seems to be wrong. I know that reaction is endothermic since the temp drops, but I am wondering which values I should be using to correctly determine the "Heat of Solution".


2 Answers 2


You have multiplied the mass of the sample, 1.50g, by temperature change and heat capacity.

However, the water provides most of the heat for the reaction.

The total mass of the solution is 1.50g + 35.0g = 36.5g.

You should be multiplying 36.5g by the temperature change and heat capacity.

Then, you need to consider how many moles 1.50g KCl is. Divide the change in enthalpy of the solution by the number of moles of KCl to determine the molar heat of solution of KCl.

  • $\begingroup$ I have done that. The answer is still wrong. If I were use 36.5g, my answer would be 0.366kj; however, my professors answer is 18.3kj $\endgroup$
    – user137452
    Dec 18, 2014 at 17:07
  • 1
    $\begingroup$ is his/her answer 18.3 kJ or 18.3 kJ/mol? $\endgroup$
    – DavePhD
    Dec 18, 2014 at 17:15
  • $\begingroup$ @user137452 if you want the answer as "per mole of KCl" you need to divide by the moles of KCl in the sample. $\endgroup$
    – DavePhD
    Dec 18, 2014 at 17:18

Here's the calculation, step by step:

$$q_\mathrm{cal} = 36.5\cdot 4.18\cdot (-2.4) = \pu{-366 J}$$ $$q_\mathrm{rxn} = -q_\mathrm{cal} = \pu{366 J}$$ $$n(\ce{KCl}) = \frac{\pu{1.50 g}}{\pu{74.55 g mol-1}} = \pu{0.0201 mol}$$
$$\frac{\pu{366 J}}{\pu{0.0201 mol}} = \pu{18.209 J mol-1} = \pu{18.2 kJ mol-1}$$

  • $\begingroup$ Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. $\endgroup$
    – andselisk
    Jan 14, 2018 at 18:52

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