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Λ⁢0𝑚⁢(NH⁢4⁢OH) is equal to ______. - Chemistry

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प्रश्न

\[\ce{\Lambda^0_m(NH4OH)}\] is equal to ______.

विकल्प

  • \[\ce{\Lambda^0_m(NH4OH) + \Lambda^0_m(NH4Cl) - \Lambda^0(HCl)}\]

  • \[\ce{\Lambda^0_m(NH4Cl) + \Lambda^0_m(NaOH) - \Lambda^0(NaCl)}\]

  • \[\ce{\Lambda^0_m(NH4Cl) + \Lambda^0_m(NaCl) - \Lambda^0(NaOH)}\]

  • \[\ce{\Lambda^0_m(NaOH) + \Lambda^0_m(NaCl) - \Lambda^0(NH4Cl)}\]

MCQ
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उत्तर

\[\ce{\Lambda^0_m(NH4OH)}\] is equal to \[\ce{\mathbf{\underline{\Lambda^0_m(NH4Cl) + \Lambda^0_m(NaOH) - \Lambda^0_m(NaCl)}}}\]

Explanation:

(i) \[\ce{NH4Cl <=> NH^{+}4 + Cl-}\]  

(ii) \[\ce{NaCl <=> Na+ + Cl-}\]  

(iii) \[\ce{NaOH <=> Na+ + OH-}\]   

(iv) \[\ce{NH4OH <=> NH^{+}4 + OH-}\] 

To get equation (iv),

\[\ce{\Lambda^0_m(NH4Cl) + \Lambda^0_m(NaOH) - \Lambda^0(NaCl) = \Lambda^0_m(NH_4OH)}\]

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अध्याय 3: Electrochemistry - Exercises [पृष्ठ ३६]

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एनसीईआरटी एक्झांप्लर Chemistry [English] Class 12
अध्याय 3 Electrochemistry
Exercises | Q I. 16. | पृष्ठ ३६

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