Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
49 views
in Chemistry by (72.2k points)
closed by
The equivalent conductance of `1 M` benzoic acid is `12.8 ohm^(-1) cm^(2)`. If the conductance of benzoate ion and `H^(+)` ion are `12` and `288.42 ohm^(-1) cm^(2)` respectively. Its degree of dissociation is :
A. `39 %`
B. `3.9 %`
C. `0.35 %`
D. `0.039 %`

1 Answer

0 votes
by (118k points)
selected by
 
Best answer
Correct Answer - B
`overset(@)(Lambda)(C_(6)H_(5)COOH) = overset(@)(Lambda)(C_(6)H_(5)COO^(-)) + overset(@)(Lambda)_((H^(+)))`
`= 42 xx 288.42 = 330.42`
`alpha = (Lambda_(m)^(C))/(overset(@)(Lambda)_(m)) = (12.8)/(330.42) = 0.0387 = 3.87%`

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...