The limiting molar conductivity $\Lambda^{\circ}$ for $NaBr$ is calculated by the following expression.
$\lambda_{N a B r}^{\infty}=\lambda_{N a C l}^{\infty}+\lambda_{K B r}^{\infty}-\lambda_{K C l}^{\infty}$
$\lambda_{N a B r}^{\infty}=126+152-150$
$=128 \,S\,cm ^{2} \,mol ^{-1}$
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(Density of water $= 1\,gm/ml$ )
$(I)$ $\begin{array}{*{20}{c}}
{C{H_2} - C{H_2} - C{H_2} - C{H_2}OH} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{N{O_2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}$
$(II)$ $\begin{array}{*{20}{c}}
{C{H_3} - C{H_2} - CH - C{H_2}OH} \\
{|\,} \\
{N{O_2}}
\end{array}$
$(III)$ $\begin{array}{*{20}{c}}
{C{H_3} - CH - C{H_2} - C{H_2}OH} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{N{O_2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}$
$(IV)$ $CH_3-CH_2-CH_2-CH_2OH$
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