A) \[201.28{{\Omega }^{-\,1}}c{{m}^{-\,2}}\]
B) \[390.71{{\Omega }^{-\,1}}c{{m}^{-\,2}}\]
C) \[698.28\,{{\Omega }^{-1}}c{{m}^{-2}}\]
D) \[540.48{{\Omega }^{-1}}c{{m}^{-2}}\]
Correct Answer: B
Solution :
Given, |
\[\lambda _{NaCl}^{\infty }=126.45\,{{\Omega }^{-1}}c{{m}^{2}}\] |
\[\lambda _{HCl}^{\infty }=426.16\,{{\Omega }^{-1}}c{{m}^{2}}\] |
\[\lambda _{{{C}_{2}}{{H}_{5}}COONa}^{\infty }=91\,{{\Omega }^{-1}}c{{m}^{2}}\] |
\[{{\lambda }_{{{C}_{2}}{{H}_{5}}COOH}}=\lambda _{{{C}_{2}}{{H}_{5}}COONa}^{\infty }+\lambda _{HCl}^{\infty }-\lambda _{NaCl}^{\infty }\] |
\[=91+4261.6-126.45\] |
\[=390.71\,{{\Omega }^{-\,1}}c{{m}^{-\,2}}\] |
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