Respuesta :
Explanation:
The given data is as follows.
For HCl : Volume = 100 mL, Density = 1.00 g/ml, Molarity = 0.25 M
For NaOH : Volume = 200 mL, Density = 1.00 g/ml, Molarity = 0.150 M
Therefore, mass of HCl will be calculated as follows.
Density = [tex]\frac{mass}{volume}[/tex]
1.00 g/ml = [tex]\frac{mass}{100 mL}[/tex]
mass = 100 g
Also, number of moles = [tex]Molarity \times Volume[/tex]
Hence, no. of moles of HCl = [tex]0.25 M \times 0.001 L[/tex] (as 1 ml = 1000 L)
= [tex]0.25 \times 10^{-3} mol[/tex]
On the other hand, mass of NaOH will be calculated as follows.
Density = [tex]\frac{mass}{volume}[/tex]
1.00 g/ml = [tex]\frac{mass}{200 mL}[/tex]
mass = 200 g
Also, number of moles = [tex]Molarity \times Volume[/tex]
Hence, no. of moles of HCl = [tex]0.150 M \times 0.002 L[/tex] (as 1 ml = 1000 L)
= [tex]0.3 \times 10^{-3} mol[/tex]
Total mass = (100 g + 200 g) = 300 g
Since, HCl is the limiting reagent over here. So, heat produced by [tex]0.25 \times 10^{-3} mol[/tex] will be calculated as follows.
Q = [tex]0.25 \times 10^{-3} mol \times \Delta H^{o}_{298}[/tex]
= [tex]0.25 \times 10^{-3} mol \times -58 kJ[/tex]
= [tex]14.5 \times 10^{-3}[/tex] kJ
= [tex]14.5 \times 10^{-3} \times 10^{3}[/tex] J
Also, it is known that relation between Q and temperature change is:
Q = [tex]mC \Delta T[/tex]
Hence, putting the values into the above formula as follows.
Q = [tex]mC \Delta T[/tex]
14.5 J = [tex]300 g \times 4.19 J/g ^{o}C \times \Delta T[/tex]
[tex]\Delta T[/tex] = [tex]0.0115 ^{o}C[/tex]
Thus, we can conclude that increase in temperature is [tex]0.0115 ^{o}C[/tex].
