A) 41.8 debye
B) 4.18 debye
C) 4.8 debye
D) 0.48 debye
Correct Answer: C
Solution :
Given ionic charge \[=4.8\times {{10}^{-10}}\] e. s. u. and ionic distance \[=1\,\overset{\text{o}}{\mathop{\text{A}}}\,={{10}^{-10}}\,\text{cm}\text{.}\] We know that dipole moment = ionic charge \[\times \] ionic distance \[=(4.8\times {{10}^{-10}})\times {{10}^{-8}}\] \[=4.8\times {{10}^{-18}}\,\]e. s. u per cm \[=4.8\,\text{debye}\text{.}\]
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