Question

If 2.00 moles of lithium bromide are dissolved in 1000.0 grams of water at 25.0 °C, what is the final temperature of the water, assuming that all solutions have the same heat capacity as pure water (4.184 J/g-K)?

Answer 1

Step-by-step explanation:

The dissociation of lithium bromide is given by the thermochemical equation as,

`LiBr\rightarrow Li^+\ +\ Br^-\ \Delta H=-48.83\ kJ/mol`

Number of moles, n = 2 moles

Molar mass of Lithium bromide is 86.845 g/mol

For the two moles, the heat released is,
`Q=2* 48.83=97.66\ kJ=97660\ J`

For 2 moles of Lithium Bromide,
`m=2* 86.845=173.69\ g`

Initial temperature,
`T_i=25^(\circ)C=298\ K`

The heat capacity is given by :

`Q=mc\Delta T`

`Q=mc(T_f-T_i)`

`T_f` is the final temperature

`T_f=(Q)/(mc)+T_i`

`T_f=(97660\ J)/(173.69\ g* 4.184\ J/g-K)+298\ K`

`T_f=432.38\ K`

So, the final temperature of the water is 432.38 k. Hence, this is the required solution.