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Chemistry Question on Conductance

The equivalent conductance of Ba2+Ba^{2+} and Cl−Cl^- are respectively 127127 and 76 ohm−1 cm2 eq−176 \,ohm^{-1}\,cm^2\, eq^{-1} at infinite dilution. What will be the equivalent conductance of BaCl2BaCl_2 at infinite dilution?

A

139.5 ohm−1 cm2 eq−1139.5\,ohm^{-1}\,cm^{2}\,eq^{-1}

B

203 ohm−1 cm2 eq−1203\,ohm^{-1}\,cm^{2}\,eq^{-1}

C

279 ohm−1 cm2 eq−1279\,ohm^{-1}\,cm^{2}\,eq^{-1}

D

101.5 ohm−1 cm2 eq−1101.5\,ohm^{-1}\,cm^{2}\,eq^{-1}

Answer

139.5 ohm−1 cm2 eq−1139.5\,ohm^{-1}\,cm^{2}\,eq^{-1}

Explanation

Solution

BaCl2→Ba2++2Cl−BaCl_{2} \to Ba^{2+}+2Cl^{-} ΛBaCl2∘=ΛBa2+∘+2ΛCl−∘\Lambda^{\circ}_{BaCl_2}=\Lambda^{\circ}_{Ba^{2+}}+2\Lambda^{\circ}_{Cl^{-}} =127+(2×76)=127+\left(2\times76\right) =279 ohm−1 cm2 eq−1=279\,ohm^{-1}\,cm^{2}\,eq^{-1} Equivalent conductivity =2792=\frac{279}{2} =139.5 ohm−1 cm2 eq−1=139.5\,ohm^{-1}\,cm^{2}\,eq^{-1}