The \(\Lambda ^o_m\) for \(CaSO_4\) at \(298 K\), if \(\Lambda ^o_{Ca^{2+}} = 119.0\, \ Scm^2mol^{-1}\) and \(\Lambda ^o_{SO_4^{2-}} = 160.0\, \ Scm^2mol^{-1}\) will be
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(279\, \ Scm^2mol^{-1}\)
By Kohlrausch’s law of independent migration of ions, the limiting molar conductivity of an electrolyte is the sum of the contributions of its ions:
\( \displaystyle \Lambda^\circ_m = \nu_+\lambda^\circ_+ + \nu_-\lambda^\circ_- \)
\( \ce{CaSO4 -> Ca^2+ + SO4^2-} \), so ν+ = ν− = 1:
\( \displaystyle \Lambda^\circ_m(\ce{CaSO4}) = \lambda^\circ(\ce{Ca^2+}) + \lambda^\circ(\ce{SO4^2-}) \)
\( \displaystyle \Lambda^\circ_m(\ce{CaSO4}) = 119.0 + 160.0 = 279.0\ \mathrm{S\,cm^2\,mol^{-1}} \)
So Λ°m for \( \ce{CaSO4} \) is 279 S cm2 mol−1.
NCERT (2026-27 edition):
Strong Electrolytes – “… The law states that limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte. Thus, if λ°Na+ and λ°Cl– are limiting molar conductivity of the sodium and chloride ions respectively, then the limiting molar conductivity for sodium chloride is given by the equation:
\( \displaystyle \Lambda^\circ_m(\ce{NaCl}) = \lambda^\circ_{\ce{Na+}} + \lambda^\circ_{\ce{Cl-}} \)
In general, if an electrolyte on dissociation gives ν+ cations and ν– anions then its limiting molar conductivity is given by:”
\( \displaystyle \Lambda^\circ_m = \nu_+\lambda^\circ_+ + \nu_-\lambda^\circ_- \)