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कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान 2nd PUC Class 12

Λ⁢0𝑚⁢(NH⁢4⁢OH) is equal to ______.

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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
रिक्त स्थान भरें
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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 Exemplar [English] Class 12
अध्याय 3 Electrochemistry
Exercises | Q I. 16. | पृष्ठ ३६

संबंधित प्रश्न

The conductivity of 0.20 mol L−1 solution of KCl is 2.48 × 10−2 S cm−1. Calculate its molar conductivity and degree of dissociation (α). Given λ0 (K+) = 73.5 S cm2 mol−1 and λ0 (C1−) = 76.5 S cm2 mol−1.


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102 × κ/S m−1 1.237 11.85 23.15 55.53 106.74

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\[\begin{array}{cc}
\end{array}\]\[\begin{bmatrix}
\ce{\Lambda^{\circ}_{H^+} = 350 S cm^2 mol^{-1}}\\
\ce{\Lambda^{\circ}_{CH_3COO^-} = 50 S cm^2 mol^{-1}}
\end{bmatrix}\]


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Discuss the variation of conductivity and molar conductivity with concentration.


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