If 8.00 g ammonium nitrate is dissolved in 1.0 l water, the water decreases in temperature from 21.0 °c to 20.39 °c. determine the molar heat of solution (in kj/mol) of the ammonium nitrate. the density of water is 1.0 g/ml and assume that the resulting solution has the same specific heat as water (4.184 j/goc).

When Heat of solution ΔH solution = m.C.ΔT
when m is the mass of solution = mass of water + mass of ammonium nitrate
mass of water = Volume * density = 1L * 1000 g/L= 1000 g
by substitution:
m of solution = 1000 g + 8 g = 1008 g
ΔT = 21°C - 20.39 °C = 0.61°C
C = 4.184 J / g.C°
by substitution in ΔH formula:
ΔH = 1008 * 4.184 * 0.61 = 2572.7 J / 1000 = 2.57KJ
∴ molar heat of solution = heat of solution / no of mol of solute
no of moles of solute (NH4NO3) = weight / molecular weight = 8g / 80 g/mol
                                                       = 0.1 mol
∴molar heat of solution        = 2.57KJ / 0.1 mol
                                          = 25.7 KJ/mol

D=m/v =57.9/3 = 19.3 g/cm^3