Question

Molar conductivity of a solution of an electrolyte $ A{{B}_{2}} $ is $ 160\text{ }oh{{m}^{-1}}c{{m}^{2}}mo{{l}^{-1}} $ . If it ionizes as $ A{{B}_{2}}\to {{A}^{+2}}+2{{B}^{-1}} $ . Its equivalent conductivity will be

• A. $ 150\text{ }oh{{m}^{-1}}c{{m}^{-1}}mo{{l}^{-1}} $
• B. $ \text{75 }oh{{m}^{-1}}c{{m}^{-1}}mo{{l}^{-1}} $
• C. $ \text{80 }oh{{m}^{-1}}c{{m}^{-1}}mo{{l}^{-1}} $
• D. all of the above

Answer 1

Answer: (C)

: Molar conductivity, $ {{\wedge }_{m}}=160\,oh{{m}^{-1}}c{{m}^{2}}/mol $ $ {{\wedge }_{m}}=\frac{k\times 1000}{Molarity} $ $ {{\wedge }_{eq}}=\frac{k\times 1000}{Normality} $ $ \frac{{{\wedge }_{eq}}}{{{\wedge }_{m}}}=\frac{Molarity}{Normality} $ $ {{\wedge }_{eq}}=\frac{{{\wedge }_{m}}\times number\text{ }of\text{ }moles\text{ }per\text{ }litre}{number\text{ }of\text{ }gm\text{ }equivalents\text{ }per\text{ }litre} $ $ {{\wedge }_{eq}}=\frac{160\times eq.\,wt}{mol.wt.} $ $ =\frac{160\times mol.\,wt.}{2\times mol.wt} $ $ =80\text{ }oh{{m}^{-1}}c{{m}^{2}}e{{q}^{-1}} $ .