Question

The molar conductivities at infinite dilution of barium chloride, sulphuric acid and hydrochloric acid are $280,860and426Sc{m}^{2}mo{l}^{-1}$ respectively. The molar conductivity at infinite dilution of barium sulphate is _____ $Sc{m}^{2}mo{l}^{-1}$. (Round off to the Nearest Integer).

Answer 1

Step 1: Given data

Molar conductivities of dilution of barium chloride, sulphuric acid and hydrochloric acid is

${\lambda }_M^{\infty }\left(BaC{l}_2\right)=280,{\lambda }_M^{\infty }\left(HCl\right)=426and{\lambda }_M^{\infty }\left(H_{2}S{O}_4\right)=860$

Step 2 : To find the molar conductivity of barium sulphate

From Kohlrausch's law

${\lambda }_{\mathrm{m}}^{\infty }\left({\mathrm{BaSO}}_4\right)={\lambda }_{\mathrm{m}}^{\infty }\left({\mathrm{Ba}}^{2+}\right)+{\lambda }_{\mathrm{m}}^{\infty }\left({\mathrm{SO}}_4^{2-}\right)$

$\begin{array}{r}{\lambda }_{\mathrm{m}}^{\infty }\left({\mathrm{BaSO}}_4\right)={\lambda }_{\mathrm{m}}^{\infty }\left({\mathrm{BaCl}}_2\right)+{\lambda }_{\mathrm{m}}^{\infty }\left({\mathrm{H}}_2{\mathrm{SO}}_4\right)\\ -2{\lambda }_{\mathrm{m}}^{\infty }\left(\mathrm{HCl}\right)\end{array}$

$=280+860-2\left(426\right)$

$=288{\mathrm{Scm}}^2{\mathrm{mol}}^{-1}$