3.15 mol of an unknown solid is placed into enough water to make 150.0 mL of solution. The solution's temperature increases by 19.2°C. Calculate ∆H for the dissolution of the unknown solid. (The specific heat of the solution is 4.18 J/g･°C and the density of the solution is 1.20 g/mL).

Enthalpy change for the dissolution of the unknown solid = 4.6 Kj/mole

Explanation:

Using Q = m x Cs x ΔT ................................(1)

where Q = Amount of heat absorbed

            m = Mass of solution

           Cs = Specific heat capacity of solution

          ΔT = Change in temperature

Given Density of solution = 1.20 g/ml

And volume of solution = 150 ml

Mass of solution = density x volume

                           = 1.2 x 150

                           = 180 g

From equation (1)

              Q = 180 x 4.18 x 19.2 = 14446.08 = 14.4 Kj

So, ΔH  of the dissolution of the unknown solid = = = 4.6 kj/mole

2b2t

explanation:

2b2t