A) 39%
B) 3.9%
C) 0.35%
D) 0.039%
Correct Answer: B
Solution :
Key Idea: \[\alpha =\frac{\lambda }{{{\lambda }_{\infty }}}\] Given, \[f_{{{C}_{6}}{{H}_{5}}CO{{O}^{-}}}^{\infty }=42\,oh{{m}^{-1}}c{{m}^{2}}\] \[f_{{{H}^{+}}}^{\infty }=288.42\,oh{{m}^{-1}}c{{m}^{2}}\] \[{{\lambda }^{\infty }}_{{{C}_{6}}{{H}_{5}}COOH={{\lambda }^{\infty }}{{C}_{6}}{{H}_{5}}CO{{O}^{-}}+{{\lambda }^{\infty }}{{H}^{+}}}\] \[=42.0+288.42\] \[=330.42\,oh{{m}^{-1}}c{{m}^{2}}\] Given, \[{{\lambda }_{v}}=12.8\,oh{{m}^{-1}}c{{m}^{2}}\] \[\alpha =\frac{{{\lambda }_{v}}}{{{\lambda }_{\infty }}}=\frac{12.8}{330.42}\] \[=0.039\] \[=0.039\times 100\] \[=3.9%\]
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