If you were to prepare a 1.00 molal solution of CaCl2 beginning with water at 27.0 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.

The dissolution of CaCl2 in water is exothermic, with deltaH soln = -81.3 kJ/mol.

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  1. Don't we need to know how much solution we are to prepare? I'll assume we are to use 1 kg water.
    1.00 m CaCl2 = 110.983 g CaCl2/kg water.
    What's the density of the solution. I will assume 1.00 g/mL which probably is not a valid assumption.
    81,300 J = 1,000g x 4.184 J/C*g x (Tfinal-27)
    Solve for Tfinal.
    Check my thinking.

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  2. 150.248

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