Electrochemistry - Result Question 4
####4. At $25^{\circ} C$ molar conductance of 0.1 molar aqueous solution of ammonium hydroxide is 9.54 $ohm^{-1} cm^{2} mol^{-1}$ and at infinite dilution its molar conductance is $238 ohm^{-1} cm^{2} mol^{-1}$. The degree or ionisation of ammonium hydroxide at the same concentration and temperature is: [NEET 2013]
(a) $20.800 %$
(b) $4.008 %$
(c) $40.800 %$
(d) $2.080 %$
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Solution:
5 .
$ \text{ (b) } \alpha=\frac{\Lambda_m}{\Lambda_m^{\infty}}=\frac{9.54}{238}=0.04008=4.008 % \text{. } $
$\stackrel{0}{\Lambda} _{m(NaOH)}=\stackrel{0}{\Lambda} _{mNa}+\stackrel{0}{\Lambda} _{mOH^{-}}$
$\therefore \stackrel{o}{\Lambda} m(NH_4^{+})+\stackrel{o}{\Lambda} m(OH^{-})$
$ -\stackrel{o}{\Lambda} _{m(NaCl)} $
