In a coffee cup calorimeter, 1.60 g NH4NO3 was mixed with 75.0 grams of water at an initial temperature of 25.00* Celsius. After the dissolution of the salt, the final temperature of the calorimeter contents was 23.34 * Celcius. Assuming the solution has a heat capacity of 4.18 J/g*C, and assuming no heat loss to the calorimeter, calculate the enthalpy of the solution for the dissolution of NH4NO3 in units of kJ/mol. Q

-26.6kJ/mol

Explanation:

The dissolution of NH₄NO₃ is:

NH₄NO₃(aq) → NH₄⁺(aq) + NO₃⁻(aq)

To solve this question we need to find the heat released in the dissolution using the equation of coffee cup calorimeter:

Q = -m*S*ΔT

Where Q is heat,

m is the mass of solution = 1.60g + 75.0g = 76.6g

S is specific heat of the calorimeter (4.18J/g°C)

And ΔT is change in temperature (25°C - 23.34°C = 1.66°C)

The heat is:

Q = -76.6g*4.18J/g°C*1.66°C

Q = -531.5J = -0.5315kJ are released

The heat released per mole = Enthalpy for the dissolution is:

Moles NH₄NO₃:

1.60g * (1mol / 80.043g) = 0.0200mol

Enthalpy for the dissolution:

-0.5315kJ / 0.0200mol =

58750.08068 is the answer i think