Question
When 3.2 g of ammonium chloride is dissolved in 75 g of water, the temperature of the solution decreases from 22.8°C to 20.1°C. What is the energy of dissolution of NH4Cl per mole? (The molecular mass of NH4Cl = 53.49 g/mol, and the specific heat of water = 4.18 J/g°C.)
A. 14.8 kJ/mol
B. 0.88 kJ/mol
C. 390 kJ/mol
D. 125 kJ/mol
Could you write he formula and plug in the values.
A. 14.8 kJ/mol
B. 0.88 kJ/mol
C. 390 kJ/mol
D. 125 kJ/mol
Could you write he formula and plug in the values.
Answers
Yes I can but I would rather you show your work and let me find the error. And please explain why you have a problem in substituting the numbers.
this is what I got. .0599*4.18*2.7=.6760
I know its not the answer so what did I do wrong
The best way to say it is that you didn't follow the script. Let's do this in two steps.
Step 1 is to calculate the heat of solution (actually cooling of the solution).
q = mass H2O x specific heat H2O x (Tfinal-Tintial)
What is q?
Step 1 is to calculate the heat of solution (actually cooling of the solution).
q = mass H2O x specific heat H2O x (Tfinal-Tintial)
What is q?
is q 846.45
Exactly. And the unit is J.
Now, that's for 0.0599 mols which you have in your first post. Convert that to J/1 mol and convert that to kJ/mol
0.84645 kJ/0.0599 mol = ? kJ/mol
Now, that's for 0.0599 mols which you have in your first post. Convert that to J/1 mol and convert that to kJ/mol
0.84645 kJ/0.0599 mol = ? kJ/mol
I get 14.1. Should I got with a.
That's what I would do.
In a earlier post you said d. Why?
That was not for this problem. I was remembering d as the correct answer for the Ni/H2O problem you worked on earlier, posted your work, and I pointed at the error you made was not substituting correctly for Tf and Ti for the Ni. As I remember Tf was 95.0 and Ti was 25.4 for a difference of about 69.6 but you had used 3.10 for that. (NOTE: these number may be slightly different since I'm trying to remember this from a couple days ago. That's a little tough to do when I work 100 or more problems a day for others. ;-).
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