Question

At ${25}^{\circ }C$, the molar conductances at infinite dilution for the strong electrolytes $NaOH,NaCl$ and $BaC{l}_2$ are $248\times {10}^{-4},126\times {10}^{-4}$ and $280\times {10}^{-4}S{m}^{2}mo{l}^{-1}$ respectively, ${\lambda }_m^{\circ }Ba(OH{)}_2$ in $S{m}^{2}mo{l}^{-1}$ is

• A. $52.4\times {10}^4$
• B. $524\times {10}^4$
• C. $4.2\times {10}^4$
• D. $262\times {10}^4$

Answer 1

Answer: (B)

$524\times {10}^4$

$BaC{l}_2+2NaOH\to Ba(OH{)}_2+2NaCl$
${\lambda }_{mBa(OH{)}_2}^{\infty }={\lambda }_{mBaC{l}_2}^{\infty }+2{\lambda }_{mNaOH}^{\infty }-2{\lambda }_{mNaCl}^{\infty }$
$=280\times {10}^{-4}+2\times 248\times {10}^{-4}-2\times 126\times {10}^{-4}$
$=(280+496-252)\times {10}^{-4}$
$=524\times {10}^{-4}S{m}^{2}mo{l}^{-1}$.