SATHEE: Electrochemistry - Result Question 4

Electrochemistry - Result Question 4

####4. At $25^{\circ} C$ molar conductance of 0.1 molar aqueous solution of ammonium hydroxide is 9.54 $o h m^{- 1} c m^{2} m o l^{- 1}$ and at infinite dilution its molar conductance is $238 o h m^{- 1} c m^{2} m o l^{- 1}$ . 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) $2.080$

Show Answer

Solution:

5 .

$(b) α = \frac{Λ_{m}}{Λ_{m}^{\infty}} = \frac{9.54}{238} = 0.04008 = 4.008$

${\overset{0}{Λ}}_{m (N a O H)} = {\overset{0}{Λ}}_{m N a} + {\overset{0}{Λ}}_{m O H^{-}}$

$∴ \overset{o}{Λ} m (N H_{4}^{+}) + \overset{o}{Λ} m (O H^{-})$

$- {\overset{o}{Λ}}_{m (N a C l)}$

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Electrochemistry - Result Question 4

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