2.0469 grams of NH4NO3 has been added to 100 ml of water. Theheat obsorbed or lo

Question

2.0469 grams of NH4NO3 has been added to 100 ml of water. Theheat obsorbed or lost by water can be calcualted using q=mCwater XT, assuming the density of water is 1g/ml, and change inT= 1.37 degrees celsius, specidif heat capacity of water=4.184J/g.k .a)calculate heat obsorbed or lost by the water in J,b)what is the molar heat of solucion in K/mol of NH4NO3. C)calucalte % error (the accepted value for heat of solution forNH4NO3 is 25.69 KJ/mol.

(for part a) I have 114.8KJ, for part b) i have 4415KJ/mol, andwhen i did part C i had a very bigg error (4415-25.69)X100%,so I amnot sure were I went wrong.)

Thank you

Explanation / Answer

A) First find the mass of the water using the density. 100 mL x 1 g/mL = 100 g Now plug this into the equation q = mcT q = (100 g)(4.184 J/g K)(1.37 K) q = 573.2 J = 0.5732 kJ B) Find the moles of NH4NO3 2.0469 g / 80 g/mol = 0.0256 mol 0.5732 kJ/0.0256 mol = 22.4 kJ/mol C) To find percent error, 25.69 - 22.4 = 0.128 = 12.8%    25.69

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