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प्रश्न
\[\ce{\Lambda^0_m(NH4OH)}\] is equal to ______.
विकल्प
\[\ce{\Lambda^0_m(NH4OH) + \Lambda^0_m(NH4Cl) - \Lambda^0(HCl)}\]
\[\ce{\Lambda^0_m(NH4Cl) + \Lambda^0_m(NaOH) - \Lambda^0(NaCl)}\]
\[\ce{\Lambda^0_m(NH4Cl) + \Lambda^0_m(NaCl) - \Lambda^0(NaOH)}\]
\[\ce{\Lambda^0_m(NaOH) + \Lambda^0_m(NaCl) - \Lambda^0(NH4Cl)}\]
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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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संबंधित प्रश्न
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(d) 400 mhos.cm2.mol-1
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\[\ce{Λ^0_m H2O}\] is equal to:
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| (ii) Li | (b) metal ion which is the weakest oxidising agent |
| (iii) Au3+ | (c) non metal which is the best oxidising agent |
| (iv) Br– | (d) unreactive metal |
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| (vii) F– | (g) metal ion which is an oxidising agent |
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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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