Equivalent conductance of 1 M propanoic acid is $10 {\mathrm{ohm}}^{-1}{\mathrm{cm}}^2{\mathrm{eq}}^{-1}$ and that at infinite dilution $200 {\mathrm{ohm}}^{-1}{\mathrm{cm}}^2{\mathrm{eq}}^{-1}$ pH of the propanoic  acid  solution is

Propanoic acid is weak acid , its degree of ionization ($\alpha$) is calculated as 

$\mathrm{\alpha } =\frac{(\mathrm{equivalent} \mathrm{conductance} \mathrm{at} \mathrm{given} \mathrm{concentration}){\wedge }_{\mathrm{e}}^{\mathrm{c}}}{(\mathrm{equivalent} \mathrm{conductance} \mathrm{at} \mathrm{infinite} \mathrm{dilution}) {\wedge }_{\mathrm{e}}^{\mathrm{o}}}=\frac{10}{200}=0.05$

$$\begin{gathered} \left[{\mathrm{H}}^{+}\right]=\mathrm{C\alpha }= 1 \times 0.05= 0.05 \\ \mathrm{C}= \mathrm{Concentration} \\ \mathrm{pH}= -\log \left[{\mathrm{H}}^{+}\right] \\ =-\log (0.05) \\ \mathrm{pH} = 1.3 \end{gathered}$$