10.8 g of solid ammonium nitrate (NH4NO3, molar mass 80.052 g/mol) is dissolved in 150 mL of water.
The temperature change, ΔT, of the system is -5.1 ºC.
Calculate the heat of solution for ammonium nitrate in kJ/mol.
2 answers
i did H=mct as .225 but thats wrong
I assume you used 10.8 g for mass but the water is what's cooled by the NH4NO3 so the water gives you the amount of cooling and that's for 10.8 g NH4NO3.
q = m*c*dT
Assuming density of water is 1 g/mL, mass H2O is 150 g.
q = 150 g x 4.184 J/g*C x -5.1 C = -3,201 kJ for 10.8 g NH4NO3.
mols NH4NO3 = 10.8 g/80.052 = 0.134 mols.
So delta H is 3201 kJ/0.135 mols or 23,711 kJ/mol NH4NO3.
You should go through and check my math. I estimated here and there and didn't use all of the decimal places.
q = m*c*dT
Assuming density of water is 1 g/mL, mass H2O is 150 g.
q = 150 g x 4.184 J/g*C x -5.1 C = -3,201 kJ for 10.8 g NH4NO3.
mols NH4NO3 = 10.8 g/80.052 = 0.134 mols.
So delta H is 3201 kJ/0.135 mols or 23,711 kJ/mol NH4NO3.
You should go through and check my math. I estimated here and there and didn't use all of the decimal places.
1 answer