Answer:
The right solution is:
(a) 2459.74 J/degree C
(b) 3271.769 KJ/moles
Explanation:
According to the question,
(a)
The heat capacity of the calorimeter will be:
= [tex]\frac{5682}{24.68-22.37}[/tex]
= [tex]\frac{5682}{2.31}[/tex]
= [tex]2459.74 \ J/degree \ C[/tex]
(b)
The change in temperature will be:
= [tex]26.77-22.37[/tex]
= [tex]4.4 \ Degree \ C[/tex]
The amount of heat released will be:
= [tex]2459.74\times 4.4[/tex]
= [tex]10822 \ Joules[/tex]
or,
= [tex]10.822 \ KJ[/tex]
Moles of benzene combusted will be:
= [tex]\frac{0.258}{78}[/tex]
= [tex]0.00330 \ Moles[/tex]
hence,
The heat combustion for 1 mol of benzene will be:
= [tex]\frac{10.822}{0.00330}[/tex]
= [tex]3271.769 \ KJ/moles[/tex]
