A mixture of  2 × 10^-3 N Na₂SO₄ and 1× 10^-2 N HCl showed a resistance of 1106 ohms using a cell of cell constant 5 cm^-1. The ionic equivalent conductance of H⁺, Cl^- are 350.0 and 76.0 ohm^-1 cm² eq^-1 respectively. Assuming all solutions to be very dilute,correct statement(s) is/are

(1) Specific conductance of the mixture

$\begin{array}{l}=\frac{\mathrm{l}}{\mathrm{aR}}=\frac{5}{1106}=4.52\times {10}^{-3}{\mathrm{Scm}}^{-1}\\ =4.52\times {10}^{-3}{\mathrm{Scm}}^{-1}\times 100{\mathrm{cmm}}^{-1}\\ =4.52\times {10}^{-1}{\mathrm{sm}}^{-1}\text{. }\end{array}$

(2) Specific conductance of HCI solution in the mixture

$\begin{array}{r}=\frac{\mathrm{C\Lambda }}{1000}=\frac{0.01\times 426}{1000}=4.26\times {10}^{-3}{\mathrm{Scm}}^{-1}\\ =4.26\times {10}^{-1}{\mathrm{Sm}}^{-1}\end{array}$

$\text{ Specific conductance due to }{\mathrm{Na}}_2{\mathrm{SO}}_4$

$=(4.52-4.26)\times {10}^{-1}=2.6\times {10}^{-2}{\mathrm{sm}}^{-1}$

(3) $2.6\times {10}^{-2}={\mathrm{C}}_{{\mathrm{Na}}^{+}}{\mathrm{\lambda }}_{{\mathrm{Na}}^{+}}+{\mathrm{C}}_{{\mathrm{SO}}_4^{2-}}{\mathrm{\lambda }}_{{\mathrm{SO}}_4^{2-}}$

$\text{ ( }c=\text{ conc. in mol }{\mathrm{m}}^{-3},\lambda ={\mathrm{ohm}}^{-1}{\mathrm{m}}^2{\mathrm{mol}}^{-1}\text{ ) }$

$\begin{array}{r}=2\times \cancel{{10}^6}\left({\mathrm{molcm}}^{-3}\right)\times 50\times {10}^{-4}\left({\mathrm{Sm}}^2{\mathrm{mol}}^{-1}\right)\times \cancel{{10}^6}{\mathrm{cm}}^3{\mathrm{mol}}^{-3}\\ +1\times {10}^{-6}\left({\mathrm{molcm}}^{-3}\right)\times \mathrm{\lambda }\\ \times {10}^6{\mathrm{cm}}^3{\mathrm{mol}}^{-3}\end{array}$

or $2.6\times {10}^{-2}=0.01+1\times {\mathrm{\lambda }}_{{\mathrm{SO}}_4^{2-}}$

or ${\lambda }_{{\mathrm{SO}}_4^{2-}}=0.026-0.01=0.016\mathrm{sm} {\mathrm{mol}}^{-1}$

(4) ${\mathrm{\Lambda }}_{0\left({\mathrm{H}}_2{\mathrm{SO}}_4\right)}={\mathrm{\Lambda }}_{0\left({\mathrm{Na}}_2{\mathrm{SO}}_4\right)}+{\mathrm{\Lambda }}_{0\left(2\mathrm{HCl}\right)}-{\mathrm{\Lambda }}_{0\left(2\mathrm{NaCl}\right)}$

$\text{ (all in molar conductances in }S{\mathrm{cm}}^2{\mathrm{mol}}^{-1}\text{ ) }$

${\begin{array}{r}=\left(2\times {\mathrm{\lambda }}_{{\mathrm{Na}}^{+}}^0+{\mathrm{\lambda }}_{{\mathrm{SO}}_4^{-}}^0\right)+2{\mathrm{\lambda }}_{{\mathrm{H}}^{+}}^0+2{\mathrm{\lambda }}_{{\mathrm{Cl}}^{-}}^0-2{\mathrm{\lambda }}_{{\mathrm{Na}}^{+}}^0-2{\mathrm{\lambda }}_{{\mathrm{Cl}}^{-}}^0\\ =260+2\times 426-2\times 126=860 {\mathrm{ohm}}^{-1}{\mathrm{cm}}^2{\mathrm{mol}}^{-1}\end{array}}^{ }$