A) 0.001
B) 0.005
C) 0.01
D) 0.02
Correct Answer: C
Solution :
Given, molar conductivity \[({{\Lambda }_{m}})=300\,{{\Omega }^{-1}}c{{m}^{2}}mo{{l}^{-1}}\] Limiting molar conductivity \[(\Lambda _{m}^{o})=150\,{{\Omega }^{-1}}c{{m}^{2}}mo{{l}^{-1}}\] Concentration = 0.02 M \[\therefore \]Degree of dissociation\[(\alpha )=\frac{{{\Lambda }_{m}}}{\Lambda _{m}^{o}}\] \[\alpha =\frac{150}{300}=0.5\] Now, by using following reaction, we can determine the value of its dissociation constant \[{{K}_{a}}=\frac{C{{\alpha }^{2}}}{(1-\alpha )}\] \[{{K}_{a}}=\frac{0.02\times {{(0.5)}^{2}}}{(1-0.5)}\] \[=\frac{0.02\times 0.5\times 0.5}{0.5}=0.01\]
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