At 18°C the mobilities of NH4+ & ClO4-ions are 6.6×10-4 cm2 volt-1 sec-1& 5.7×10-4cm2 volt-1sec-1at infinite dilution. Calculate equivalent conductance of ammonium chlorate solution

λNH4+=UNH4+×F=6.6×10-4×96,500=63 .69λClO4-=UClO4-×F=5.7×10-4×96,500=55.005λNH4ClO4=63.69+55.005=118.695

At 18°C the mobilities of NH4+ & ClO4-ions are 6.6×10-4 cm2 volt-1 sec-1& 5.7×10-4cm2 volt-1sec-1at infinite dilution. Calculate equivalent conductance of ammonium chlorate solution

λNH4+=UNH4+×F=6.6×10-4×96,500=63 .69λClO4-=UClO4-×F=5.7×10-4×96,500=55.005λNH4ClO4=63.69+55.005=118.695

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At $18 ° C$ the mobilities of ${NH}_{4}^{+}$ & ${ClO}_{4}^{-}$ ions are $6.6 \times 10^{- 4} {cm}^{2} {volt}^{- 1} \sec^{- 1} & 5.7 \times 10^{- 4} {cm}^{2} {volt}^{- 1} \sec^{- 1}$ at infinite dilution. Calculate equivalent conductance of ammonium chlorate solution

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118.67

34.26

49.14

57.26

answer is A.

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Detailed Solution

$λ_{{NH}_{4}^{+}} = U_{{NH}_{4}^{+}} \times F$

$= 6.6 \times 10^{- 4} \times 96, 500$

=63 .69

$λ_{{ClO}_{4}^{-}} = U_{{ClO}_{4}^{-}} \times F$

$= 5.7 \times 10^{- 4} \times 96, 500$

$= 55.005$

$λ_{{NH}_{4} {ClO}_{4}} = 63.69 + 55.005$

=118.695

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At 18°C the mobilities of NH4+ & ClO4-ions are 6.6×10-4 cm2 volt-1 sec-1& 5.7×10-4cm2 volt-1sec-1at infinite dilution. Calculate equivalent conductance of ammonium chlorate solution

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At $18 ° C$ the mobilities of ${NH}_{4}^{+}$ & ${ClO}_{4}^{-}$ ions are $6.6 \times 10^{- 4} {cm}^{2} {volt}^{- 1} \sec^{- 1} & 5.7 \times 10^{- 4} {cm}^{2} {volt}^{- 1} \sec^{- 1}$ at infinite dilution. Calculate equivalent conductance of ammonium chlorate solution