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Electrochemistry question

2012 · Q100
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Electrochemistry question

2012 · Q100

NEETChemistryElectrochemistryMCQ+4 / −1
Limiting molar conductivity of NH4OH
[\left[ {} \right.[i.e.  Λm(NH4OH)0\Lambda _{m\left( {N{H_4}OH} \right)}^0Λm(NH4​OH)0​]\left. {} \right]] is equal to
  1. A
    Λm(NH4OH)0+Λm(NaCl)0−Λm(NaOH)0\Lambda _{m\left( {N{H_4}OH} \right)}^0 + \Lambda _{m\left( {NaCl} \right)}^0 - \Lambda _{m\left( {NaOH} \right)}^0Λm(NH4​OH)0​+Λm(NaCl)0​−Λm(NaOH)0​
  2. B
    Λm(NaOH)0+Λm(NaCl)0−Λm(NH4Cl)0\Lambda _{m\left( {NaOH} \right)}^0 + \Lambda _{m\left( {NaCl} \right)}^0 - \Lambda _{m\left( {N{H_4}Cl} \right)}^0Λm(NaOH)0​+Λm(NaCl)0​−Λm(NH4​Cl)0​
  3. C
    Λm(NH4OH)0+Λm(NH4Cl)0−Λm(HCl)0\Lambda _{m\left( {N{H_4}OH} \right)}^0 + \Lambda _{m\left( {N{H_4}Cl} \right)}^0 - \Lambda _{m\left( {HCl} \right)}^0Λm(NH4​OH)0​+Λm(NH4​Cl)0​−Λm(HCl)0​
  4. D
    Λm(NH4Cl)0+Λm(NaOH)0−Λm(NaCl)0\Lambda _{m\left( {N{H_4}Cl} \right)}^0 + \Lambda _{m\left( {NaOH} \right)}^0 - \Lambda _{m\left( {NaCl} \right)}^0Λm(NH4​Cl)0​+Λm(NaOH)0​−Λm(NaCl)0​
View written solutionFree

Correct answer: D

According to Kohlrausch’s law, the molar conductivity of NH4OH

Λm(NH4OH)0\Lambda _{m\left( {N{H_4}OH} \right)}^0Λm(NH4​OH)0​ = Λm(NH4Cl)0+Λm(NaOH)0−Λm(NaCl)0\Lambda _{m\left( {N{H_4}Cl} \right)}^0 + \Lambda _{m\left( {NaOH} \right)}^0 - \Lambda _{m\left( {NaCl} \right)}^0Λm(NH4​Cl)0​+Λm(NaOH)0​−Λm(NaCl)0​

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