How would I calculate the heat of solution in Joules for dissolving 0.1 mole of NH4NO3 in 100 mL of water?
The initial temperature was 22.8 degrees celsius and the temperature after the salt dissolved was 18.1 degrees celsius.
Dissolving NH4NO3 is an endothermic reaction. How much heat is absorbed can be measured by using the temperature of the water.
q = mass H2O x specific heat H2O x (Tfinal-Tinitial).
That is q for 0.1M so
q/0.1 = q in kJ/mol.
Do I have to add q of the caloriemeter to that?
If there is a calorimeter constant you didn't type the entire question and my response will not be right.. You may need to include. Henceforth, please don't edit the question.
To calculate the heat of solution of NH4NO3 in Joules, you need to use the formula:
Heat of Solution = mass of solute × specific heat capacity of water × change in temperature
1. First, you need to calculate the mass of NH4NO3 used. This can be done using the molar mass of NH4NO3, which is 80.043 g/mol.
Mass of NH4NO3 = number of moles × molar mass
= 0.1 mol × 80.043 g/mol
= 8.0043 g
2. Next, you need to convert the volume of water from milliliters (mL) to grams (g) as the specific heat capacity of water is given in J/g°C.
Density of water = 1 g/mL
Mass of water = volume of water × density of water
= 100 mL × 1 g/mL
= 100 g
3. Now, you can calculate the change in temperature by subtracting the initial temperature from the final temperature.
Change in temperature = final temperature - initial temperature
= 18.1°C - 22.8°C
= -4.7°C
4. Finally, you can calculate the heat of solution using the formula mentioned earlier.
Heat of Solution = mass of NH4NO3 × specific heat capacity of water × change in temperature
= 8.0043 g × 4.184 J/g°C × (-4.7°C)
≈ -157.06 J
The negative sign indicates that the dissolution of NH4NO3 is an exothermic process, meaning heat is released during dissolution. Therefore, the heat of solution in this case is approximately -157.06 Joules.