Question

The molar conductivities of $AgN{O}_3,NaCl$ and $NaN{O}_3$ at infinite dilution are 116.0, 110.0 and $105.0Sc{m}^{2}mo{l}^{–1}$ , respectively. The conductivity of $AgCl$ in water is $8.42\times {10}^{-6}Sc{m}^{–1}$ and of water used is $1.16\times {10}^{–6}Sc{m}^{-1}$. Find the solubility of AgCl in $mollitr{e}^{–1}$

• A. $7.92\times {10}^{-5}$
• B. $9.58\times {10}^{-6}$
  
• C. $6.96\times {10}^{-5}$
  
• D. $6.0\times {10}^{-5}$

Answer 1

The correct option is D $6.0\times {10}^{-5}$

${\lambda }_{AgCl}^0={\lambda }_{AgN{O}_3}^0+{\lambda }_{NaCl}^0-{\lambda }_{NaN{O}_3}^0=116+110-105=121Sc{m}^{2}mo{l}^{-1}$

Conductivity due to AgCl without $H_{2}O=8.42\times {10}^{-6}-1.16\times {10}^{-6}=7.26\times {10}^{-6}Sc{m}^{-1}$

${\lambda }_{AgCl}^0=\frac{k\times 1000}{s}=\frac{7.26\times {10}^{-6}\times 1000}{s}=121\Rightarrow s=\frac{7.26\times {10}^{-3}}{121}=6.0\times {10}^{-5}M$