Given the following molar conductivities at 25⁰ C; HCl, $426{\mathrm{\Omega }}^{-1}{\mathrm{cm}}^2{\mathrm{mol}}^{-1}; \mathrm{NaCl},126{\mathrm{\Omega }}^{-1}{\mathrm{cm}}^2{\mathrm{mol}}^{-1}$; NaC (sodium crotonate) $83{\mathrm{\Omega }}^{-1}{\mathrm{cm}}^2{\mathrm{mol}}^{-1}, \mathrm{x}\times {10}^{-5}$ is the ionization constant of crotonic acid, if the conductivity of a 0.001 M crotonic acid solution is $3.83\times {10}^{-5}{\mathrm{\Omega }}^{-1}{\mathrm{cm}}^{-1}$. x =