Question

The heat capacity of a certain calorimeter is 2.30 kJ·K^–1 when containing 1.00 L of water. When 10.0 g of ammonium nitrate are dissolved in water to make 1.00 L of solution in this calorimeter, the temperature drops 1.52 K. What is the molar enthalpy of solution of ammonium nitrate?

I know the answer is +28.0 kJ·mol^–1, but can someone explain the steps, please? Thank you.

Answer 1

the heat gained or lost can be calculated using the formula
Q = mc(T2 - T1)
where
m = mass (grams)
c = specific heat capacity (J/g-K)
T = temperature (K)

the specific heat of the calorimeter, which is c = 2.3 J/g-K. (Please note that you have been provided 2.30 kJ·K^–1, which shoud be 2.30 J·K^–1) - I mean it is in Joules NOT in kilo joules)
The total mass of solution is 1000 + 10 = 1010 grams. density of water = 1 g/mL, so 1 .00 L = 1000 mL = 1000 g
Substituting,
Q = (1010 g)*(2.3 J/g.K)*(1.52 K)
Q = +3530.96 J (Sign should be positive because of endothermic reaction)

ammonium nitrate molar mass = 80.043 g/mol

moles = mass of ammonium nitrate / molar mass of ammonium nitrate
moles in 10 g ammonium nitrate = 10g / 80.043 g/mol = 0.12493284859 mol

molar enthalpy of solution = energy / moles ammonium nitrate = Q / mol

molar enthalpy of solution = 3530.96 J / 0.12493284859 mol = 28263.5075642 J/mol = ~ 28 kJ.mol-1

the molar enthalpy of solution of ammonium nitrate = +28.0 kJ·mol^–1

Hope this helped you!

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