I have already tried the answers: -29.42, +29.42, and -10.579. The standard enth

Question

I have already tried the answers: -29.42, +29.42, and -10.579.

The standard enthalpy change for the dissolution of ammonium nitrate is shown below. If 14.0 grams of solid ammonium nitrated is dissolved in 100.0 grams of water at a temperature of 20.0 degree C, what will be the final temperature of the solution? You may assume that the specific heat of the solution is the same as the specific heat of pure water (4.184 J/g- degree C) and the the calorimeter itself absorbes no heat. NH4NO3(S) rightarrow NH4(aq) + NO3(aq) Delta H degree = +25.69 KJ Final temperature of the solution:

Explanation / Answer

moles of NH4NO3(s) = g/molar mass = 14 g/80.05 g/mol = 0.175 mol

q = delta H = mCp(Tf-Ti) this is for 1 mole

So, for 0.175 moles

delta H/0.175 = mCp(Tf-Ti)

where,

delta H = 25.69 kJ = 25690 J

m = total mass = 14 + 100 = 114 g

Cp = 4.184 J/g.oC

Tf = final temperature of solution

Ti = initial temperature of solution

Feed values,

25.69/0.175 = 114 x 4.184 x (Tf-20)

Tf = 20.31 oC is the final temperature of solution

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