In a coffee-cup calorimeter experiment, 10.00 g of a soluble ionic compound was added to the calorimeter containing 75.0 g h2o initially at 23.2°c. the final temperature of the solution was 31.8°c. what was the change in enthalpy for the dissolution of this compound? give your answer in units of joules per gram of compound. assume that the specific heat of the solution is the same as that of pure water, 4.18 j ⁄ g ⋅ °c. a) 2.7 x 10^2 j/g b) -3.1 x 10^3 j/g c) -3.1 x 10^2 j/g d) 3.1 x 10^2 j/g

Given:

Mass of the ionic compound  = 10.00 g

Mass of water = 75.0 g

Initial temperature of water T1= 23.2 C

Final temperature of water T2 = 31.8 C

Specific heat of water c = 4.18 J/gC

To determine:

Enthalpy of dissolution of the ionic compound

Explanation:

Heat gained by water is given as:

Q = mcΔT

m = mass of water

c = specific heat

ΔT = change in temp of water = T2-T1

Q = 75.0 g * 4.18 J/gC * (31.8-23.2)C = 2696 J

Now,

Heat gained by water = heat lost by the ionic compound (i.e. heat of dissolution)

Thus, q(ionic) = 2696 J

ΔH = q(ionic)/mass of ionic compound = 2696 J/10.00 g = 2.7 *10² J/g

Ans: A) enthalpy change = 2.7*10² J/g