Question

When 2.35g Mg(OH)2 is added to 250.0 mL of water, the temperature of the water raises from 20.5 oC to 36.0 oC. Calculate the molar enthalpy of solution.

Answer 1

Amount of energy absorbed by water:

$q=cm\Delta T = 4.186 \frac{J}{g\times ^\circ C} \times 250.0 g \times (36.0 ^\circ C-20.5 ^\circ C)= 16220.75 J$

Moles of solute:

$2.35 g Mg(OH)_2 \times (\frac{1 mole Mg(OH)_2}{58.32 g Mg(OH)_2})= 0.0403 mol$

As the temperature is increased then the reaction is exothermic, then:

$\Delta H_{soln}= \frac{-q}{n}$

$\Delta H _{soln} = \frac{-16220.75 J}{0.0403 mol}=-402500 \frac{J}{mol}=- 403 \frac{kJ}{mol}$