Question

Heat of solution of NH₄Cl

A. Determine the number of calories and joules absorbed when the NH4Cl dissolved.

B. Determine the number of mole of NH4Cl dissolved.

C. Determine the heat of solution of NH4Cl in cal/g, Kcal/mole, J/g and kJ/mole.

Heat of solution of NH4Cl

Initial water temperature (Trial 1 &2) = 20°C

Volume of water in cup (Trial 1 &2) = 50 ml

Mass of NH₄Cl (Trial 1 &2) = 5.198 g, 5.005g

Final Temperature of Solution (Trial 1&2) = 13°C, 14°C

Total Mass of Solution (Trial 1&2) = 54.335 g, 53.168g

Temperature change (Trial 1&2) = 7°C, 6°C

Answer 1

The initial temperature = 20°C

Volume of water = 50mL

Trial 1

Final temperature = 13 °C

So change in temprature = 20-13 = 7°C

The heat absorbed = Mass of water X change in temprature x specific heat of water

Heat absrobed = 50 X 7 X 4.18 = 1463 Joules

B) Moles of NH4Cl dissolved = Mass of NH4Cl / Molecular weight = 5.198 / 53.5 = 0.097 moles

C) Heat absorbed of NH4Cl = Heat lost by water = 1463 Joules

(i) 1463 J by 0.097moles = 1.463 KJ / 0.097 = 15.08 KJ / moles

(ii) 1463 J / 5.198 grams = 281.45 Joules / gram

(iii) 1 joule = 0.239 Calories

1463 Joules = 349.65 calories

349.65 calories / 5.198 grams = 67.27 cal / g

(iv) 0.349Kcal / 0.097 moles =3.597 Kcal / mole

Trial 2

Final temperature = 14 °C

So change in temprature = 20-14= 6°C

The heat absorbed = Mass of water X change in temprature x specific heat of water

Heat absrobed = 50 X 6 X 4.18 = 1254 Joules

B ) Moles of NH4Cl dissolved = Mass of NH4Cl / Molecular weight = 5.005 / 53.5 = 0.093 moles

C) Heat absorbed of NH4Cl = Heat lost by water = 1254 Joules

(i) 1254 J by 0.093 moles = 1.254 KJ / 0.093 = 13.48 KJ / moles

(ii) 1254 J / 5.005 grams = 250.55 Joules / gram

(iii) 1 joule = 0.239 Calories

1254 Joules = 299.7 calories

299.7 calories / 5.005 grams = 59.88 cal / g

(iv) 0.299 Kcal / 0.093 moles = 3.215 Kcal / mole