Question

From the given molar conductivities infinite dilution, calculate ${\lambda }_m^{\infty }$ for $N{H}_{4}OH$.
${\lambda }_m^{\infty }\text{for}Ba(OH{)}_2=457.6oh{m}^{-1}c{m}^{2}mo{l}^{-1}.$
${\lambda }_m^{\infty }\text{for}Ba\left(C{l}_2\right)=240.6oh{m}^{-1}c{m}^{2}mo{l}^{-1}.$
${\lambda }_m^{\infty }\text{for}N{H}_{4}Cl=128.8oh{m}^{-1}c{m}^{2}mo{l}^{-1}.$

• A. $-238.3oh{m}^{-1}c{m}^{2}mo{l}^{-1}$
• B. $212.6oh{m}^{-1}c{m}^{2}mo{l}^{-1}$
• C. $338.4oh{m}^{-1}c{m}^{2}mo{l}^{-1}$
• D. $238.3oh{m}^{-1}c{m}^{2}mo{l}^{-1}$

Answer 1

The correct option is D $238.3oh{m}^{-1}c{m}^{2}mo{l}^{-1}$

${\lambda }_m^{\infty }\text{for}Ba(OH{)}_2={\lambda }_{B{a}^{2+}}^0+2{\lambda }_{O{H}^{-}}^0....(i)$
${\lambda }_m^{\infty }\text{for}\left(BaC{l}_2\right)={\lambda }_{B{a}^{2+}}^0+2{\lambda }_{C{l}^{-}}^0....\left(ii\right)$
${\lambda }_m^{\infty }\text{for}\left(N{H}_{4}Cl\right)={\lambda }_{N{H}_4^{+}}^0+{\lambda }_{C{l}^{-}}^0....\left(iii\right)$
$\because {\lambda }_m^{\infty }f\left(N{H}_{4}OH\right)={\lambda }_{N{H}_4^{+}}^0+{\lambda }_{O{H}^{-}}^0$
= 1/2 eq.(i) + eq.(iii)-1/2 eq.(ii)
$=\frac{1}{2}\times 457.6+129.8-\frac{1}{2}\times 240.6$
$=238.3oh{m}^{-1}c{m}^{2}mo{l}^{-1}$