Respuesta :
Explanation:
The given data is as follows.
[tex]V_{1}[/tex] = 85.0 ml, [tex]M_{1}[/tex] = 0.9 M
[tex]V_{2}[/tex] = 85.0 ml, [tex]M_{1}[/tex] = 0.9 M
Hence, number of moles of HCl and KOH will be the same because both the solutions have same volume and molarity.
So, No. of moles = Molarity × Volume
= [tex]0.9 M \times 0.085 L[/tex] (as 1 L = 1000 ml so, 85 ml = 0.085 L)
= 0.076 mol
As 1 mole gives 56.2 kJ/mol of heat of neutralization. Hence, calculate the heat of neutralization given by 0.076 moles as follows.
[tex]56.2 kJ/mol \times 0.076 mol[/tex]
= 4.271 kJ
or, = 4271 J (as 1 kJ = 1000 J)
Therefore, heat released = - heat of gained by calorimeter
Since, it is given that density of the solution is similar to the density of water which is 1 g/ml.
Hence, mass of HCl = density × Volume of HCl
= 1.00 g/ml × 85.0 ml
= 85 g
Similarly, mass of KOH = = density × Volume of HCl
= 1.00 g/ml × 85.0 ml
= 85 g
Hence, total mass of the solution = 85 g + 85 g
= 170 g
Also, q = [tex]mC \Delta T[/tex]
4271 J = [tex]170 g \times 325 J/^{o}C \times (T_{f} - 18.24)^{o}C[/tex]
0.0773 = [tex]T_{f} - 18.24[/tex]
[tex]T_{f}[/tex] = [tex]18.317^{o}C[/tex]
Thus, we can conclude that final temperature of the mixed solution is [tex]18.317^{o}C[/tex].
