Explanation:
The given data is as follows.
Total mass of KBr solution (m) = 125 g + 10.5 g
= 135.5 g
Change in temperature (dT) = 24.2 - 21.1 = [tex]3.1^{o}C[/tex]
Specific heat (C) = 4.184 [tex]J/g^{o}C[/tex]
Hence, heat absorbed will be calculated as follows.
q = mC dT
= [tex]135.5 g \times 4.184 J/g^{o}C \times 3.1^{o}C[/tex]
= 1757.5 J
Therefore, enthalpy change per gram of salt will be calculated as follows.
[tex]\frac{1757.5 J}{135.5 g}[/tex]
= 12.97 J/g
So, no. of moles present in 10.5 g of KBr are calculated as follows.
No. of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]
= [tex]\frac{10.5 g}{119 g/mol}[/tex]
= 0.0882 mol
Change in enthalpy will be calculated as follows.
[tex]\frac{1757.5 kJ}{1000 \times 0.0882 mol}[/tex]
= 19.93 kJ/mol
Thus, we can conclude that the enthalpy change for dissolving the given salt is 19.93 kJ/mol and 12.97 J/g.