Question

Specific conductance of a saturated solution of $AgBr$ is $8.486\times {10}^{-7}$ $oh{m}^{-1}$ $c{m}^{-1}$ at ${25}^o$C. Specific conductance of pure water at ${25}^o$C is $0.75\times {10}^{-6}$ $oh{m}^{-1}$ $c{m}^{-2}$. ${\Lambda }_m^{\infty }$ for $KBr$, $AgN{O}_3$ and $KN{O}_3$ are $137.4,133,131$(S $c{m}^2$ $mo{l}^{-1}$) respectively. The solubility of $AgBr$ is $4\times {10}^{-4}$gm/litre.

• A. True
• B. False

Answer 1

The correct option is B False

S.conductance of $AgBr=8.486\times {10}^{-7}on{m}^{-1}c{m}^{-1}$

S.conductance of $H_{2}O=7.5\times {10}^{-7}on{m}^{-1}c{m}^{-1}$

${\lambda }_m^{\infty }\left(AgBr\right)={\lambda }_m^{\infty }\left(KBr\right)+{\lambda }_m^{\infty }(AgN{o}_3-{\lambda }_m^{\infty }KN{O}_3)$

${\lambda }_m^{\infty }\left(AgBr\right)=137.4+133-131=139.4sc{m}^{2}mo{l}^{-1}$

${\lambda }_m^{\infty }=\frac{K\times 1000}{m\left(ors\right)}$ (s: solubility of saturated solution)

$139.4=\left(\frac{KAgBr-K{H}_{2}0\times 1000}{5}\right)$

$139.4=\frac{(8.486-7.5)\times {10}^{-7}}{5}\times 1000$

$S\left(AgBr\right)=707\times {10}^{-7}gm/Litre$

So answer is false