Question

The molar conductivity at infinite dilution of $AgN{O}_3,NaCl$ and $NaN{O}_3$ are $116.5$, $110.3$ and $105.2$ S $c{m}^{2}mo{l}^{-1}$ respectively. The conductivity of $AgCl$ in water is $2.40\times {10}^{-6}Sc{m}^{-1}$ and of water used is $1.16\times {10}^{-6}Sc{m}^{-1}$. The solubility of $AgCl$ will be:

• A. $1.463\times {10}^{-3}g-litr{e}^{-1}$
• B. $1.163\times {10}^{-2}g-litr{e}^{-1}$
• C. $1.163\times {10}^{-3}g-litr{e}^{-1}$
• D. None of these

Answer 1

The correct option is A $1.463\times {10}^{-3}g-litr{e}^{-1}$

For monovalent electrolytes,

Equivalent conductivity $=$ Molar conductivity

Given, $A_{AgN{O}_3}^{\infty }=116.5={\lambda }_{A{g}^{+}}^{\infty }+{\lambda }_{N{O}_3^{-}}^{\infty }$ ......(i)

$A_{NaCl}^{\infty }=110.3={\lambda }_{N{a}^{+}}^{\infty }+{\lambda }_{C{l}^{-}}^{\infty }$ ......(ii)

$A_{NaN{O}_3}^{\infty }=105.2={\lambda }_{N{a}^{+}}^{\infty }+{\lambda }_{N{O}_3^{-}}^{\infty }$ ......(iii)

$\therefore A_{AgCl}^{\infty }=Eqs.\left[\right(i)+(ii)-(iii\left)\right]$

$=116.5+110.3-105.2=121.6$
For sparingly soluble salt ${\lambda }_v={\lambda }^{\infty }$

$\therefore A_{vAgCl}=121.6Sc{m}^{2}e{q}^{-1}$

Now, $k_{AgCl+water}=2.40\times {10}^{-6}Sc{m}^{-1}$

$k_{water}=1.16\times {10}^{-6}Sc{m}^{-1}$

$\therefore k_{AgCl}=(2.40-1.16)\times {10}^{-6}$

$=1.24\times {10}^{-6}Sc{m}^{-1}$

Therefore, $A_v=k\times \left(1000/M\right)(\because N=M)$

$121.6=1.24\times {10}^{-6}\times \left(1000/M\right)$

$\therefore M=\frac{1.24\times {10}^{-3}}{121.6}mol/litre$

$c=\frac{1.24\times {10}^{-3}}{121.6}\times 143.5g/litre$

Solubility of AgCl $=1.463\times {10}^{-3}glitr{e}^{-1}$