If you were to prepare a 1.00 m solution of CaCl2 beginning with water at 29.5∘C, what would the final tempera?
The dissolution of CaCl2(s) in water is exothermic, with ΔHsoln = -81.3 kJ/mol. If you were to prepare a 1.00 m solution of CaCl2 beginning with water at 29.5∘C, what would the final temperature of the solution be (in ∘C)? Assume that the specific heats of both pure H2O and the solution are the same, 4.18 J/(K⋅g).
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Basis - 1 lit 1 molar CaCl2 soln that is 110.98 gm CaCl2 dissolved in 1 lit water
Molecular mass of CaCl2 is 110.98 g/mol
Mass of soln = (110.98 + 1000) = 1110.98gm
sp heat of soln = 4.186 joule/gram °C
Delta T = (X - 29.5) deg c , let X is the final temperature in degc
Q ( heat liberated ) = 81.3 * 1000 j
Using standard equation of heat transfer
Q( heat) = mass*sp heat*delta T
(X-29.5) =( 81.3*1000) / (1110.98*4.186)= 17.5 degc
therefore, X = 29.5+17.5 = 47 degc