Question

The order of equivalent conductances at infinite dilution of $LiCl,NaCl$ and $KCl$ is:

• A. $LiCl>NaCl>KCl$
• B. $KCl>NaCl>LiCl$
• C. $NaCl>KCl>LiCl$
• D. $LiCl>KCl>NaCl$

Answer 1

The correct option is B $KCl>NaCl>LiCl$

The decreasing order of equivalent conductance at infinite dilution of $LiCl,NaCl$ and $KCl$ is $KCl>NaCl>LiCl.$

The increasing order of ionic radii is:

$L{i}^{+}\left(aq\right)<N{a}^{+}\left(aq\right)<K^{+}\left(aq\right)$

The decreasing order of hydrated radii is:

$L{i}^{+}\left(aq\right)>N{a}^{+}\left(aq\right)>K^{+}\left(aq\right)$

The increasing order of ionic mobility is:

$L{i}^{+}\left(aq\right)<N{a}^{+}\left(aq\right)<K^{+}\left(aq\right)$