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(D) The molar conductivity at infinite dilution is given by $\wedge_{m}^{\infty} = x + y \ S \ cm^2 \ mol^{-1}$.The relationship between specific cond

Vedclass · Accepted Answer

$\frac{\kappa 	imes 188 	imes 1000}{x + y}$

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(D) The molar conductivity at infinite dilution is given by $\wedge_{m}^{\infty} = x + y \ S \ cm^2 \ mol^{-1}$.The relationship between specific conductance $\kappa$,molar conductivity $\wedge_{m}$,and molarity $M$ (which is equal to solubility $S$ for a saturated solution) is $\wedge_{m}^{\infty} = \frac{\kappa 	imes 1000}{S}$.Rearranging for solubility $S$ in $mol/L$: $S = \frac{\kappa 	imes 1000}{x + y} \ mol/L$.To convert solubility from $mol/L$ to $g/L$,multiply by the molar mass of $AgBr$ $(108 + 80 = 188 \ g/mol)$:$S (g/L) = \frac{\kappa 	imes 1000 	imes 188}{x + y} \ g/L$.

The specific conductance of a saturated solution of silver bromide is $\kappa \ S \ cm^{-1}$. The limiting ionic conductances for $Ag^{+}$ and $Br^{-}$ are $x$ and $y$ $S \ cm^2 \ mol^{-1}$ respectively. Then the solubility of silver bromide (in $g/L$) will be $(Ag = 108, Br = 80)$.

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