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The dissolution of $\mathrm{CaCl}_{2}(s)$ in water is exothermic, with $\Delta H_{\mathrm{soln}}=-81.3 \mathrm{kJ} / \mathrm{mol}$ . If you were to prepare a 1.00 $\mathrm{m}$solution of $\mathrm{CaCl}_{2}$ beginning with water at $25.0^{\circ} \mathrm{C},$ whatwould the final temperature of the solution be in $^{\circ} \mathrm{C} ?$Assume that the specific heats of both pure $\mathrm{H}_{2} \mathrm{O}$ and thesolution are the same, 4.18 $\mathrm{J} /(\mathrm{K} \cdot \mathrm{g})$ .
Final temperature of solution $=44.4^{\circ} \mathrm{C}$
Chemistry 102
Chapter 11
Solutions and Their Properties
Solutions
University of Maryland - University College
Brown University
University of Toronto
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in order to solve for the temperature change of the dissolution of a one moral solution of calcium chloride, we first have to recognize that one more well is equivalent to one more of calcium chloride, where every one kilogram of water from here we have what we need to solve the equation. Q equals M C Delta T to figure out the change in temperature, we start with Q. Which is the change in panto P for the solution multiplied by the number of moles. Since we're given the Moller and will be of solution, and then we could set this equal to emcee Dr T. For every one mole of calcium chloride, 81,300 jewels are produced. End of the 1000 grams the water. The specific heat is 4.18 Jules her Calvin Graham. From here, we can solve for the change in temperature as 19 4 Kelvin's or 19.4 degrees Celsius. This means the final temperature is just 25 degrees plus 19 degrees, and it comes out to 44.4 degrees Celsius.
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