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

Answer 1

Equating the heat released from enthalpy change to the heating of the solution (assume c_Δsoln = c_W of water):

-nΔH_Soln = m_solnc_wΔT where n is the moles of Mg(OH)₂ and the mass of the solution is 255 g and c_w=4.187 J/g°C.

This can now be solved for ΔH_soln. If this is an intro chemistry class, the compound is an unfortunate choice as it does not dissolve very well in water. If this is a more advanced class you would need to calculate the equilibrium concentration of Mg²⁺ to do this problem. I get that about .0016 grams of Mg(OH)₂ dissolves. This would mean that the heat of solution would be much larger than before as you divide by the number of moles of the compound that actually dissolves. If x is the molarity of Mg²⁺, then

5.61 x 10^-12 = x*(2x)² at room temp