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jee main offline problem Resistance of 0.2 M solution of an electrolyte is 50 ohm. The specific conductance of the solution is 1.4 S m-1. The resistance of 0.5 M solution of the same electrolyte is 280ohm. The molar conductivity of 0.5 M solution of the electrolyte in S m2 mol-1 is: (1) 5 × 103 (2) 5 × 102 (3) 5 × 10-4 (4) 5 × 10-3 i answered 5*10^2 which contradicts with both fiitjee and resonance keys(5*10^-4 is given as answer) kA/L=1/R LET L/A=X K/x=1/R 1.4*50=X=70m-1 MOLAR CONDUCTIVITY(in si units)=k/c(in si units) m.c.=X/RC-----(C-MOL/m3) MC=70/(280*0.5*10^-3) MC=0.25*2*10^3=5*10^2 IN SI UNITS could any one debug this please

jee main offline problem
Resistance of 0.2 M solution of an electrolyte is 50 ohm. The specific conductance of the solution is 1.4 S m-1.
The resistance of 0.5 M solution of the same electrolyte is 280ohm. The molar conductivity of 0.5 M solution
of the electrolyte in S m2 mol-1 is:
(1) 5 × 103 (2) 5 × 102
(3) 5 × 10-4 (4) 5 × 10-3
i answered 5*10^2 which contradicts with both fiitjee and resonance keys(5*10^-4 is given as answer)

kA/L=1/R
LET L/A=X
K/x=1/R
1.4*50=X=70m-1

MOLAR CONDUCTIVITY(in si units)=k/c(in si units)
m.c.=X/RC-----(C-MOL/m3)
MC=70/(280*0.5*10^-3)
MC=0.25*2*10^3=5*10^2 IN SI UNITS
could any one debug this please

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1 Answers

Sunil Kumar FP
askIITians Faculty 183 Points
10 years ago
you have done correct but in the last molar conductivity=.25*2/10^3=5*10^-4
because 1M=10^-3mol/m^3 it is divided by not multiplication.

thanks and regards
sunil kumar
askIItian faculty

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