Solveeit Logo
Login

Question

Chemistry Question on Conductance

The limiting molar conductivities Λ∘\Lambda^{\circ} for NaCl, KBr and KCl are 126, 152 and 150 S cm2 mol−1cm^2\, mol^{-1} respectively. The Λ∘\Lambda^{\circ} for NaBr is

A

128 S cm2 mol−1128\, S \,cm^2\, mol^{-1}

B

302 S cm2 mol−1302\, S \,cm^2\, mol^{-1}

C

278 S cm2 mol−1278\, S \,cm^2\, mol^{-1}

D

176 S cm2 mol−1176\, S \,cm^2\, mol^{-1}

Answer

128 S cm2 mol−1128\, S \,cm^2\, mol^{-1}

Explanation

Solution

ΛNaCl∘=λNa∘+λCl∘=126…(1)\Lambda^{\circ}_{NaCl} = \lambda^{\circ}_{Na} +\lambda^{\circ}_{Cl} = 126\quad\dots \left(1\right) ?KBr∘=?K+∘+?Br−∘=152…(2)?^{\circ}_{KBr} = ?^{\circ}_{K^{+}} + ?^{\circ}_{Br^{-}} = 152\quad\quad\dots \left(2\right) ?KCl∘=?K+∘+?Br−∘=150…(3)?^{\circ}_{KCl} = ?^{\circ}_{K^{+}} + ?^{\circ}_{Br^{-}} = 150\quad\quad\dots \left(3\right) ΛNaBr∘=?Na∘+?Br−∘\Lambda^{\circ}_{NaBr} = ?^{\circ}_{Na} + ?^{\circ}_{Br^{-}} ΛNaBr∘=126+152−150=128\Lambda ^{\circ }_{NaBr} = 126 + 152 -150 = 128