Answer to Question #131796 in General Chemistry for Lorence Delos Santos

Question

The heat of a solution of ammonium chloride is 15.2kJ/mol. If a 6.134g sample of NH4Cl is added to 65.0mL of water in a calorimeter at 24.5C, what is the minimum temperature reached by the solution? (The specific heat of water = 4.18J/gC; the heat capacity of the calorimeter = 365J/C

Answer 1

Moles of NH4Cl:

6.134 g / 53.491 g/mol = 0.1147 mol NH4Cl

Heat absorbed

 0.1147 mol X 15.2 kJ/mol = 1.743 kJ

Volume of the water = 65.0 mL = 0.065 L

Mass of the water = 1000 g/L × 0.065 L = 65 g

Total mass = 65 g + 6.134 g = 71.134 g

Q = m*c*(T2-T1)

-1743 J = 71.134 g (4.18 J/gC) (T2-24.5)

T2 = 18.6 C