student Jorge performs an experiment in which he adds 11.40 g KNO3 to 63.80 mL o

Question

student Jorge performs an experiment in which he adds 11.40 g KNO3 to 63.80 mL of deionized water in a coffee cup calorimeter. He calculates the enthalpy change of the solution for the process to be +1818.31 J. This value can be expressed as the molar enthalpy change of dissolution for KNO3, in kJ/mol. What is Jorge's molar enthalpy change of dissolution (kJ/mol)? Express your response to two digits after the decimal.

------------------------------

student Sarah is measuring the heat of dissolution of KNO3 in water. She adds 2.81 g of solid KNO3 to 41.75 mL of deionized water in a coffee cup calorimeter (mass of the coffee cups + the stir bar = 20.32 g). The initial temperature of the water is 56.33 oC, and the temperature of the solution after the dissolution process is complete is 11.52 oC. Assuming the specific heat of the solution is the same as that of pure water (4.184 J/g-K), what is the heat of solution for the dissolution process in J (Joules)?

Explanation / Answer

HRxn = 1818.31

MW of KNO3 = 101.1032

mol = mass/MW = 11.40/101.1032 = 0.112756 mol of KNO3

HRxn = Q/mol = (1818.31)/0.112756 = 16126.059 J/mol = 16.12 kJ/mol

Q2.

Qsolution = m*C*(Tf-Ti)

Qsolution = (41.75+2.81)(4.184)*(11.52-56.33)

mol of KNO3 = masS/MW = 2.851/101.1032 = 0.0281989

So

HRxn = -Q/n = --8354.3/ 0.0281989 = 296263.3294 J

HRx = 296263.3294/1000 kJ = 296.2633 kJ/mol

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.