following data was collected after running a calorimetric experiment using three
ID: 878196 • Letter: F
Question
following data was collected after running a calorimetric experiment using three Table 1: AHsol Values of Salts Salt | Hsol (kJ/mol) LiI -63.60 -2.51 4.73 LiNO3 LiF The data collected for each experiment is as follows Experiment 1: 2.91 g of unknown salt added to 180, mL of water resulted in a value of-1.8°C. Experiment 2: 2.91 g of unknown salt added to 180. mL of water resulted in a value of-0.1°C. Experiment 3: 2.91 g of unknown salt added to 180. mL of water resulted in a T value of 0.7°C. Calculate the value of q for each of the experiments. (Assume that the heat capacity of water is 4.1801 J/g.°c.) Experiment 1 Experiment 2 4. Experiment 3 4 ent 2 9Explanation / Answer
Solution :-
Lets calculate the energy absorbed in the experiment 1
Mass of salt = 2.91 g
Mass of water = 180 g
Change in temperature = -1.8 oC
Formula
q=m*s*delta T
where q= energy, m= mass , s= specific heat and delta T = change on temperature.
Lets put the values in the formula.
q=m*s*delta T
= 180g *4.184 J per g.oC * 1.8 oC
= 1356 J
1356 J*1 kJ / 1000 J = 1.356 kJ
Now lets calculate it for peer mol of each salt
For LiI ,
1.356 kJ * 133.841 g per mol / 2.91 g = 62.4 kJ/ mol
It is close to given value of the LiI that 63.6 kJ per mol
There fore its LiI salt used in experiment 1
Lets calculate the energy absorbed in the experiment 2
Mass of salt = 2.91 g
Mass of water = 180 g
Change in temperature = -0.1 oC
Formula
q=m*s*delta T
where q= energy, m= mass , s= specific heat and delta T = change on temperature.
Lets put the values in the formula.
q=m*s*delta T
= 180g *4.184 J per g.oC *0.1 oC
= 75.3 J
75.3 J*1 kJ / 1000 J = 0.0753 kJ
Now lets calculate it for peer mol of each salt
Lets try for LiNO3 ,
0.0753 kJ * 68.946 g per mol / 2.91 g = 1.78 kJ/ mol
It is close to given value of the LiNO3 that is 2.51 kJ per mol
There fore its LiNO3 salt used in experiment 2
Lets calculate the energy absorbed in the experiment 3
Mass of salt = 2.91 g
Mass of water = 180 g
Change in temperature = 0.7 oC
Formula
q=m*s*delta T
where q= energy, m= mass , s= specific heat and delta T = change on temperature.
Lets put the values in the formula.
q=m*s*delta T
= 180g *4.184 J per g.oC *0.7 oC
= 527.2 J
527.2 J*1 kJ / 1000 J = 0.5272 kJ
Now lets calculate it for peer mol of each salt
Lets try for LiF ,
0.5272 kJ * 25.939g per mol / 2.91 g = 4.70kJ/ mol
It is close to given value of the LiF that is 4.73 kJ per mol
There fore its LiF salt used in experiment 3
There Q values for each experiment are as follows
Experiment 1 = q = -1356 J
Experiment 2 = q = -75.3 J
Experiment 3 = q = 527.2 J
Identity of the salt in each experiment is as follows.
Experiment 1 = LiI
Experiment 2 = LiNO3
Experiment 3= LiF
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