Question

The specific conductance of a saturated solution of silver bromide is $\kappa \text{S}\text{c}{\text{m}}^{-1}$. The limiting ionic conductivity of $\text{A}{\text{g}}^{+}\text{and}\text{B}{\text{r}}^{-}$ ions are x and y, respectively. The solubility of silver bromide in $\text{g}{\text{L}}^{\text{- 1}}$ is: [molar mass of AgBr = 188]

• A. $\frac{\kappa \times 1000}{\text{x}-\text{y}}$
• B. $\frac{\kappa }{\text{x}+\text{y}}\times 188$
• C. $\frac{\kappa \times 1000\times 188}{\text{x}+\text{y}}$
• D. $\frac{\text{x}\times \text{y}}{\kappa }\times \frac{1000}{188}$

Answer 1

The correct option is C $\frac{\kappa \times 1000\times 188}{\text{x}+\text{y}}$

Given Specific conductance of AgBr = K S $\text{c}{\text{m}}^{-1}$

$\Lambda {}_{A{g}^{+}}^{\circ }=x$ $\Lambda {}_{B{r}^{-}}^{\circ }=y$

$\Lambda {}_{AgBr}^{\circ }=\Lambda {}_{A{g}^{+}}^{\circ }+\Lambda {}_{B{r}^{-}}^{\circ }=x+y$

$\Lambda {}_{AgBr}^{\circ }=\text{K}\times \frac{1000}{\text{C}\left(\text{mol}{\text{L}}^{-1}\right)}$

$x+y=\frac{\text{K}\times 1000}{\text{C}\left(\text{mol}{\text{L}}^{-1}\right)}$

(Solubility) C = $\left(\frac{K\times 1000}{x+y}\right)\text{mol}.{\text{L}}^{-1}$

No. of mol = $\frac{\text{mass(g)}}{\text{molar}\text{mass}\left(\text{g}\text{mo}{\text{l}}^{\text{- 1}}\right)}$

molar mass of AgBr = 188 $\text{g}\text{mo}{\text{l}}^{\text{- 1}}$
So, $\text{C}=\left(\frac{\text{K}\times 1000}{x+y}\right)\times 188\text{g}\text{.}{\text{L}}^{-1}$

Hence, option (C) is correct.