0.700 moles of an unknown solid is placed into water to make 150.0 mL of solution. The solution's temperature decreases by 6.11 °C. Calculate ∆H, in kJ/mol, for the dissolution of the unknown solid.

Answer 1

q = mC∆T

q = heat = ?

m = mass of water = 150.0 ml x 1 g / ml = 150.0 g (assuming a density of 1.0 g / ml)

C = specific heat of water = 4.184 J/gº

∆T = change in temperature = 6.11º

Solving for q, we have...

q = (150.0 g)(4.184 J/gº)(6.11º) = 3834.6 J

∆H in kJ = 3834.6 J x 1 kJ / 1000 J = 3.835 kJ

∆H in kJ/mol = 3.835 kJ / 0.700 mols = 5.48 kJ/mol (3 sig. figs.)

NOTE: since the temperature of the solution decreased, the process is endothermic, and hence the sign of ∆H is positive (+5.48 kJ/mole)

Answer 2

The enthalpy of dissolution (dHdiss) will be given by the heat of dissolution divided by 0.700 moles. To find qdiss we will use

qdiss = -qwater

= -mH2OxCsxdT

= -(150.0g)x(4.184 J/gC)x(6.11 C) .

I assumed density and specific heat of solution are that of water.

qdiss = +3832 J = 3.832 kJ

dHdiss = 3.832 kJ/.700 mol= +5..47 kJ/mole