Question

At $25{}^{\circ }\text{C}$, in one litre saturated solution of silver oxalate $(K_{sp}=1.6\times {10}^{-8})$, 0.1 moles of $N{a}_2{C}_2{O}_4$ are added. If ${\lambda }_{\left(A{g}^{+}\right)}^{\infty }=60{\text{S cm}}^2{\text{mol}}^{-1}$, ${\lambda }_{\left(C_2{O}_4^{2-}\right)}^{\infty }=140{\text{S cm}}^2{\text{mol}}^{-1}$, ${\lambda }_{\left(N{a}^{+}\right)}^{\infty }=50{\text{S cm}}^2{\text{mol}}^{-1}$ then select the correct statement(s).

• A. Conductance of the solution is nearly $24\times {10}^{-6}\text{S}$.
• B. Conductivity of the solution is nearly $24\times {10}^{-3}{\text{S cm}}^{-1}$
• C. If ${\lambda }_{\left(C{l}^{-}\right)}^{\infty }$ is $50{\text{S cm}}^2{\text{mol}}^{-1}$ then ${\lambda }_{\left(AgCl\right)}^{\infty }=110{\text{S cm}}^2{\text{mol}}^{-1}$.
• D. Solubility of $A{g}_2{C}_2{O}_4$ in the presence of $N{a}_2{C}_2{O}_4$ is $2\times {10}^{-4}\text{mol/L}$.

Answer 1

Answer: (B), (C), (D)

The correct options are
B Conductivity of the solution is nearly $24\times {10}^{-3}{\text{S cm}}^{-1}$
C If ${\lambda }_{\left(C{l}^{-}\right)}^{\infty }$ is $50{\text{S cm}}^2{\text{mol}}^{-1}$ then ${\lambda }_{\left(AgCl\right)}^{\infty }=110{\text{S cm}}^2{\text{mol}}^{-1}$.
D Solubility of $A{g}_2{C}_2{O}_4$ in the presence of $N{a}_2{C}_2{O}_4$ is $2\times {10}^{-4}\text{mol/L}$.

$A{g}_2{C}_2{O}_4\text{(s)}\rightleftharpoons 2A{g}^{+}\text{(aq)}+C_2{O}_4^{2-}\text{(aq)}$
$2s$ $\left(0.1\right)\text{M}$
$K_{sp}={\left[A{g}^{+}\right]}^2{\left[C_2{O}_4^{2-}\right]}^1$
$1.6\times {10}^{-8}=\left(2s{)}^2\right(0.1{)}^1$
$s=2\times {10}^{-4}\text{M}$
So,
$K_{\text{solution}}=(K{)}_{A{g}^{+}}+(K{)}_{C_2{O}_4^{2-}}+(K{)}_{N{a}^{+}}$
$=\frac{60\times (4\times {10}^{-4})}{1000}+\frac{140\times 0.1}{1000}+\frac{50\times 0.2}{1000}$
$\simeq 24\times {10}^{-3}{\text{S cm}}^{-1}$
Also,
${\lambda }_M^{\infty }\left(AgCl\right)={\lambda }_M^{\infty }\left(A{g}^{+}\right)+{\lambda }_M^{\infty }\left(C{l}^{-}\right)$
${\lambda }_M^{\infty }\left(AgCl\right)=60+50=110{\text{S cm}}^2{\text{mol}}^{-1}$