A) 0.75
B) 0.25
C) 0.125
D) 0.50
Correct Answer: C
Solution :
Given,\[\Lambda _{m}^{o}NaCl=126.4S\,c{{m}^{2}}mo{{l}^{-1}}\] \[\Lambda _{m}^{o}HCl=425.9S\,c{{m}^{2}}mo{{l}^{-1}}\] \[\Lambda _{m}^{{}}NaA=100.5S\,c{{m}^{2}}\,mo{{l}^{-1}}\] \[\Lambda _{m}^{o}HA=\Lambda _{m}^{o}HCl+\Lambda _{m}^{o}NaA-\Lambda _{m}^{o}NaCl\] \[=425.9+100.5-126.4\] \[=400S\,c{{m}^{2}}mo{{l}^{-1}}\] \[\Lambda _{m}^{{}}\frac{1000K}{C}=5\times {{10}^{-5}}\times \frac{1000}{0.001}=50\] \[\alpha =\frac{\Lambda _{m}^{{}}}{\Lambda _{m}^{o}}=\frac{50}{400}=0.125\]You need to login to perform this action.
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