Question

The limiting molar conductivities of the given electrolytes at $298$K follow the order: $\left[{\lambda }^0\right(K^{+})=73.5,{\lambda }^0(C{l}^{-})=76.3,{\lambda }^0(C{a}^{2+})=119.0,{\lambda }^0(S{O}_4^{2-})=160.0S$ $c{m}^{2}mo{l}^{-1}]$.

• A. $KCl<CaC{l}_2<K_{2}S{O}_4$
• B. $KCl<K_{2}S{O}_4<CaC{l}_2$
• C. $K_{2}S{O}_4<CaC{l}_2<KCl$
• D. $CaC{l}_2<K_{2}S{O}_4<KCl$

Answer 1

Answer: (A)

$KCl<CaC{l}_2<K_{2}S{O}_4$

From Kohlrausch's law,

Molar conductivity of $KCl={\lambda }^o\left(K^{+}\right)+{\lambda }^o\left(C{l}^{-}\right)=73.5+76.3=149.8$

Molar conductivity of $CaC{l}_2={\lambda }^o\left(C{a}^{2+}\right)+2\times {\lambda }^o\left(C{l}^{-}\right)=119+(2\times 76.3)=271.6$

Molar conductivity of $K_{2}S{O}_4=2\times {\lambda }^o\left(K^{+}\right)+{\lambda }^o\left(S{O}_4^{2-}\right)=2\times 73.5+160=307$

So, limiting molar conductivity of electrolytes follow the order: $KCl<CaC{l}_2<K_{2}S{O}_4$.