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Chemistry Question on Electrochemistry

λm∘\lambda^{\circ}_m for NH4Cl,NaOH {NH_4Cl, NaOH} and NaCl {NaCl} are 130,248130, 248 and 126.5 ohm−1 cm2 mol−1 {126.5\, ohm^{-1} \, cm^2 \, mol^{-1}} respectively. The λm∘\lambda^{\circ}_m of NH4OH {NH_4OH}

A

251.5251.5

B

244.5244.5

C

130130

D

504.5504.5

Answer

251.5251.5

Explanation

Solution

λm(NH4OH)∘=λm(NH4Cl)∘+λm(NaOH)∘−λm(NHCl)∘\lambda^{\circ}_{m( {NH_4OH}) } = \lambda^{\circ}_{m( {NH_4Cl}) } + \lambda^{\circ}_{m( {NaOH}) } - \lambda^{\circ}_{m( {NHCl}) }
=130+248−126.5=251.5 ohm−1 cm2 mol−1{ = 130 + 248 - 126.5 = 251.5 \, ohm^{-1} \, cm^{2} \, mol^{-1} }