Question

A potential difference of 20 volts applied to the ends of a column of M/10 AgNO solution. 4 cm in
diameter and 12 cm in length gave a current of 0.20 A. Calculate the specific and molar conductivities
of the solution


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Answer 1

Answer:

By ohm's law, V=IR

R=  20 /20 = 1Ω

Area of section =πr*2

=3.14×2×2=12.56 $$cm^2

Length of cell 12 cm

Conductivity =  

1 /R×1/A  =  1 /1  ×  12.56 /12

=0.9554 Scm  −1

 Molar Conductivity = K/C  ×1000

=  0.9554/0.1   × 1000

=9554 cm 2 Ωmol  −1

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