ABC26PH0058 · Electrochemistry

Subject: Physical Chemistry · Chapter: Electrochemistry · Topic: Conductance · Exam: CSIR-NET DEC 2013 · Marks: 2 · Difficulty: Medium

The limiting molar conductivities of $\mathrm{NaCl}$, $\mathrm{NaI}$ and $\mathrm{RbI}$ are $12.7$, $10.8$ and $9.1\,\mathrm{mS\,m^{2}\,mol^{-1}}$, respectively. The limiting molar conductivity of $\mathrm{RbCl}$ would be
(a)$32.6\,\mathrm{mS\,m^{2}\,mol^{-1}}$
(b)$7.2\,\mathrm{mS\,m^{2}\,mol^{-1}}$
(c)$14.4\,\mathrm{mS\,m^{2}\,mol^{-1}}$
(d)$11.0\,\mathrm{mS\,m^{2}\,mol^{-1}}$
Answer
Answer (as printed): D
Explanation
$\lambda^{\infty}_{\mathrm{NaCl}}=\lambda^{\infty}_{\mathrm{Na^{+}}}+\lambda^{\infty}_{\mathrm{Cl^{-}}}\ \ldots(1)$; $\lambda^{\infty}_{\mathrm{NaI}}=\lambda^{\infty}_{\mathrm{Na^{+}}}+\lambda^{\infty}_{\mathrm{I^{-}}}\ \ldots(2)$; $\lambda^{\infty}_{\mathrm{RbI}}=\lambda^{\infty}_{\mathrm{Rb^{+}}}+\lambda^{\infty}_{\mathrm{I^{-}}}\ \ldots(3)$. Add (1) and (3), subtract (2): $\lambda^{\infty}_{\mathrm{Na^{+}}}+\lambda^{\infty}_{\mathrm{Cl^{-}}}+\lambda^{\infty}_{\mathrm{Rb^{+}}}+\lambda^{\infty}_{\mathrm{I^{-}}}-\lambda^{\infty}_{\mathrm{Na^{+}}}-\lambda^{\infty}_{\mathrm{I^{-}}}=12.7+9.1-10.8$. $\lambda^{\infty}_{\mathrm{Cl^{-}}}+\lambda^{\infty}_{\mathrm{Rb^{+}}}=21.8-10.8$. $\lambda^{\infty}_{\mathrm{RbCl}}=11.0\,\mathrm{mS\,m^{2}\,mol^{-1}}$.

Open in whiteboard · Browse this chapter in the app