Find heat of solution, delta hs, of ammonia chloride in kj/mol. This is a lab so

Question

Find heat of solution, delta hs, of ammonia chloride in kj/mol.
This is a lab so I'm not entirely sure what's needed.
3.467 grams of NH4Cl was placed into a calorimeter containing 100 mL DI water that was at 20.6 degrees C. The temperature dropped to 18 degrees C.
The calorimeter constant is 150.9 j/c. I found the dT to be -2.6 degrees C, the heat flow of the solution to be -.0359 kj, the heat flow of the calorimeter to be -.3923 kj, and the total dH for dissolution to be .428 kj. Find heat of solution, delta hs, of ammonia chloride in kj/mol.
This is a lab so I'm not entirely sure what's needed.
3.467 grams of NH4Cl was placed into a calorimeter containing 100 mL DI water that was at 20.6 degrees C. The temperature dropped to 18 degrees C.
The calorimeter constant is 150.9 j/c. I found the dT to be -2.6 degrees C, the heat flow of the solution to be -.0359 kj, the heat flow of the calorimeter to be -.3923 kj, and the total dH for dissolution to be .428 kj.
This is a lab so I'm not entirely sure what's needed.
3.467 grams of NH4Cl was placed into a calorimeter containing 100 mL DI water that was at 20.6 degrees C. The temperature dropped to 18 degrees C.
The calorimeter constant is 150.9 j/c. I found the dT to be -2.6 degrees C, the heat flow of the solution to be -.0359 kj, the heat flow of the calorimeter to be -.3923 kj, and the total dH for dissolution to be .428 kj.

Explanation / Answer

HN4Cl(s) + H2O --- > NH4Cl(aq), DeltaH = + Ve

Since the temperature of the calorimeter and water is decreased, heat is absorbed during the dossolution of NH4Cl(s) in water. Hence DeltaH for the above reaction is +ve.

Mass of NH4Cl(s) taken = 3.467 g

Molecular mass of NH4Cl = 53.49 g/mol

Hence moles of NH4Cl(s) taken = mass / molecular mass = 3.467 g / 53.49g/mol = 0.06482 mol

Heat released by water = mxsxdT = 100g x (4.184J/g.DegC) x (- 2.6 DegC) = - 1088 J

Heat released by calorimeter = CxdT = 150.9 J/DegC x (- 2.6 DegC) = - 392.3 J

Hence total heat released by the system = - (1088 + 392.3) J = -1480.3 J

Now heat absorbed during the dissolution of 0.06482 mol NH4Cl in above reaction = - ( -1480.3 J) = +1480.3 J

Hence heat absorbed during the dissolution of 1 mol NH4Cl = (+1480.3 J) / 0.06482 mol = 22840 J/mol

= 22.84 KJ/mol

Hence heat of solution of NH4Cl = 22.84 KJ/mol (answer)

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