JEE Main 12 Jan 2019 Evening Question 26
$ \Lambda {{{}^\circ } _{m}} $ for $ NaCl,HCl $ and NaA are 126.4, 425.9 and $ 100.5,S,cm^{2},mol^{-1}, $ respectively. If the conductivity of 0.001 M HA is $ 5\times {10^{-5}}S,cm^{-1}, $ degree of dissociation of HA is [JEE Main Online Paper Held On 12-Jan-2019 Evening]
Options:
A) 0.75
B) 0.25
C) 0.125
D) 0.50
Show Answer
Answer:
Correct Answer: C
Solution:
- Given, $ \Lambda _m^{o}NaCl=126.4,S,cm^{2},mol^{-1} $
$ \Lambda _m^{o} , HCl=425.9,S,cm^{2}mol^{-1}$
$ \Lambda _m^{{}} ,NaA=100.5,S,cm^{2},mol^{-1}$
$ \Lambda _m^{o} HA=\Lambda _m^{o}HCl+\Lambda _m^{o}NaA-\Lambda _m^{o}NaCl $
$ =425.9+100.5-126.4 $ $ =400S,cm^{2},mol^{-1} $ $ \Lambda _m^{{}}\frac{1000}{C}=5\times {10^{-5}}\times \frac{1000}{0.001}=50 $
$ \alpha =\frac{\Lambda _m^{{}}}{\Lambda _m^{o}}=\frac{50}{400}=0.125 $
BETA
