What mass (in grams) of CaCl2(s) must be dissolved in pure water at 10.0 deg Celsius to make a 26.4 mL solution and to increase the solution temperature to 16.4 deg Celsius? Assume that there is no heat loss from the solution and that the solution has the same physical properties as pure water.

Question

Data:
DHsol(CaCl2)(s)) = - 82.8 kJ.mol-1

specific heat of water = 4.184 J.g-1.deg-1

density of water = 1.00 g.cm-3

molar mass of CaCl2 = 111 g/mol

Answer 1

(26.4mL)(1.00g/mL) = 26.4g H2O

To raise the temperature of 26.4 g H2O by (16.4 C- 10.0 C), we need:
(4.18 J/g.C)(26.4 g)(6.4 C) = 706 J needed

- 82.8 kJ.mol-1 CaCl2 is equivalent to:
82800 J / 111g CaCl2 = 746 J/g CaCl2 released.

Use the values: 706 J needed and 746 J/g CaCl2 released to get the grams of CaCl2 that must be used.

Answer 2

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