A) \[Xe{{O}_{3}}\]
B) \[Xe{{O}_{4}}\]
C) \[3:5\]
D) \[3:5\]
Correct Answer: B
Solution :
\[BaC{{l}_{2}}+2NaOH\xrightarrow{{}}Ba{{(OH)}_{2}}+2NaCl\] \[\lambda {{_{m}^{\infty }}_{Ba{{(OH)}_{2}}}}=\lambda {{_{m}^{\infty }}_{BaC{{l}_{2}}}}+2\lambda _{m\,\,NaOH}^{\infty }-2\lambda _{m\,\,NaCl}^{\infty }\] \[=280\times {{10}^{-4}}+2\times 248\times {{10}^{-4}}\] \[-2\times 126\times {{10}^{-4}}\] \[=(280+496-252)\times {{10}^{-4}}\] \[=524\times {{10}^{-4}}S{{m}^{2}}\,mo{{l}^{-1}}\]You need to login to perform this action.
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