Question

The limiting equivalent conductivity of $NaCl,KCl\text{and}KBr$ are $126.5,150.0$ and $151.5Sc{m}^{2}e{q}^{-1}$, respectively .The limiting equivalent ionic conductivity for $Br$ is $78Sc{m}^{2}e{q}^{-1}$ .The limiting equivalent ionic conductivity for $N{a}^{+}$ ions would be :

• A. $128$
• B. $125$
• C. $50$
• D. $49$

Answer 1

The correct option is C $50$

Correct Answer : option D
${\Lambda }_m^{\infty }\left(NaBr\right)={\Lambda }_m^{\infty }\left(NaCl\right)+{\Lambda }_m^{\infty }KBr-{\Lambda }_m^{\infty }\left(KCl\right){\lambda }_m^{\infty }\left(N{a}^{+}\right)+{\lambda }_m^{\infty }\left(B{r}^{-}\right)$
$=126.5+151.5-150$
${\lambda }_m^{\infty }\left(N{a}^{+}\right)=50Sc{m}^{2}e{q}^{-1}$