JEE Advanced Model Paper 5 with solutions for online practice

Answer (A, C, D)
R = 50 Ω
C =

1

50

s
Cell constant = l/a
=

2

3

cm−1
= 66.67 m−1
Specific conductance , K → kappa = 66.67 ×

1

50

= 1.33 Sm−1
Concentration = 0.5 mol/lit = 0.5 mol/dm3
= 500 mol/m3
Molar conductivity =

K

C

=

1.33

500

2.66× 10−3Sm2mol−1