Question

Specific conductance of $0.1M$ nitric acid is $6.3\times {10}^{-2}oh{m}^{-1}c{m}^{-1}$. The molar conductance of solution is

• A. $630oh{m}^{-1}c{m}^{2}mo{l}^{-1}$
• B. $315oh{m}^{-1}c{m}^{2}mo{l}^{-1}$
• C. $100oh{m}^{-1}c{m}^{2}mo{l}^{-1}$
• D. $1201oh{m}^{-1}c{m}^{2}mo{l}^{-1}$

Answer 1

The correct option is A

$630oh{m}^{-1}c{m}^{2}mo{l}^{-1}$

$\text{Molar concentration,}\left(C\right)=0.1M$
$\text{Conductivity,}\kappa =6.3\times {10}^{-2}oh{m}^{-1}c{m}^{-1}$
$\text{Molar conductance,}{\Lambda }_m=\frac{K}{C}$
Also, we know
$1L=1000c{m}^3$
$1L^{-1}={10}^{-3}c{m}^{-3}$
$\text{Molar concentration,}\left(C\right)=0.1mol{L}^{-1}$
$\text{Molar concentration,}\left(C\right)=0.1\times {10}^{-3}molc{m}^{-3}$

$\text{Molar conductance,}{\Lambda }_m=\frac{6.3\times {10}^{-2}}{0.1\times {10}^{-3}}$
$\text{Molar conductance,}{\Lambda }_m=6.3\times {10}^{2}oh{m}^{-1}c{m}^{2}mo{l}^{-1}$