JEE Main 2019 — Electrochemistry Question with Solution
From: JEE Main 2019 (Online) 12th January Evening Slot
Question
for NaCl, HCl and NaA are 126.4, 425.9 and 100.5 S cm2 mol–1, respectively. If the conductivity of 0.001 M HA is-
5 10–5 S cm–1, degree of dissociation of HA is -
5 10–5 S cm–1, degree of dissociation of HA is -
Choose an option
Show full solutionCorrect option: B
Correct answer
B0.125
Step-by-step explanation
(HA) = (HCl) + (NaA) (NaCl)
= 425.9 + 100.5 126.4
= 400 S cm2 . mol1
=
=
= 50 S cm2 mol1
= = = 0.125
= 425.9 + 100.5 126.4
= 400 S cm2 . mol1
=
=
= 50 S cm2 mol1
= = = 0.125
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