Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6
Banner 7
Banner 8
Banner 9

Q.

In the electrochemical conversion (Kolbe’s eletrolysis) of R–COONa to R–R, 1A current was passed for 965 seconds. Calculate the amount of R–R formed in this process (Faraday constant = 96,500 C mol–1)

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.
An Intiative by Sri Chaitanya

a

10m mol

b

5m mol

c

100m mol

d

50m mol

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

Detailed Solution

2RCOONa+2H2O⟶2FR- Anode R+2CO2+2NaOH+H2 Cathode 1 mole of R-R = 2 F = 2 x 96500 coulombNo.of Coulombs consumed = i×t =1 x 965 =965 C No.of moles of R-R formed =9652×96500=1200 mol  ; Number of milli moles=5 ;
Watch 3-min video & get full concept clarity

courses

No courses found

score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

whats app icon