Explanation:
The dissociation of lithium bromide is given by the thermochemical equation as,
[tex]LiBr\rightarrow Li^+\ +\ Br^-\ \Delta H=-48.83\ kJ/mol[/tex]
Number of moles, n = 2 moles
Molar mass of Lithium bromide is 86.845 g/mol
For the two moles, the heat released is, [tex]Q=2\times 48.83=97.66\ kJ=97660\ J[/tex]
For 2 moles of Lithium Bromide, [tex]m=2\times 86.845=173.69\ g[/tex]
Initial temperature, [tex]T_i=25^{\circ}C=298\ K[/tex]
The heat capacity is given by :
[tex]Q=mc\Delta T[/tex]
[tex]Q=mc(T_f-T_i)[/tex]
[tex]T_f[/tex] is the final temperature
[tex]T_f=\dfrac{Q}{mc}+T_i[/tex]
[tex]T_f=\dfrac{97660\ J}{173.69\ g\times 4.184\ J/g-K}+298\ K[/tex]
[tex]T_f=432.38\ K[/tex]
So, the final temperature of the water is 432.38 k. Hence, this is the required solution.
