Question

The molar conductivity of a $0.5mol/d{m}^3$ solution of $AgN{O}_3$ with electrolytic conductivity of $5.76\times {10}^{-3}Sc{m}^{-1}$ at 298 K is

• A. $0.086Sc{m}^2/mol$
• B. $28.8Sc{m}^2/mol$
• C. $2.88Sc{m}^2/mol$
• D. $11.52Sc{m}^2/mol$

Answer 1

The correct option is D $11.52Sc{m}^2/mol$

$C=0.5mol/d{m}^3$
$k=5.76\times {10}^{-3}Sc{m}^{-1}$
What this indicates is that if we have a cell of 1 mL electrolyte solution, its conductivity will be equal to $k$.

$T=298K$

${\lambda }_m$ is conductivity of 1 mole of the electrolyte. That is why to obtain it we divide $k$ by concentration to obtain the value for 1 mole and multiply it with 1000 (to extrapolate to 1 L from 1 mL).

${\lambda }_m=\frac{k\times 1000}{M}$
$=\frac{5.76\times {10}^{-3}\times {10}^{+3}}{0.5}=11.52Sc{m}^2/mol$