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)?
Endothermic
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".