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The molar solubility is 7.4×[tex]10^{-3}[/tex] M and the solubility is 7.4×[tex]10^{-3}[/tex] g/L .
Calculation ,
The dissociation of silver bromide is given as ,
[tex]AgBr[/tex] → [tex]Ag ^{+}[/tex] + [tex]Br^{-}[/tex]
S
- S S
Ksp = [[tex]Ag ^{+}[/tex] ] [ [tex]Br^{-}[/tex] ] = [S] [ S ] = [tex]S^{2}[/tex]
S = √ Ksp = √ 5. 5×[tex]10^{-5}[/tex] = 7.4×[tex]10^{-3}[/tex]
The solubility =7.4×[tex]10^{-3}[/tex] g/L
The molar solubility is the solubility of one mole of the substance.
Since , one mole of [tex]AgBr[/tex] is dissociates and form one mole of each [tex]Ag ^{+}[/tex] and [tex]Br^{-}[/tex] ion . So, solubility is equal to molar solubility but unit is different.
Molar solubility = 7.4×[tex]10^{-3}[/tex] mol/L = 7.4×[tex]10^{-3}[/tex] M
To learn more about molar solubility ,
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