Question

5. The electrical resistance of 0.08 M caustic soda solution is 5 x 103 when placed in a column of diameter 2 cm and length 50 cm. The conductivity of the solution is: (Take the value of n = 3) s: A. O 1/300 S cm 1 O -1 B. O 300 S cm C. O 300 S cm -1 D. O1/300 S cm​

Answer 1

Answer:

(a) The resistance of M/10 solution is found to be 2.5×10

−3

ohms. Calculate the molar conductance. (Given cell contant =1.15cm

−1

).

Medium

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Answer 2

Answer: $1/300Scm^{-1}$ is the conductivity of the given solution

Step-by-step explanation:

Given:

$R=5\times 10^{3}\\d=2cm\\l=50cm\\Area=\pi r^{2}=3.14\times 1\times 1=3.14\\R=\rho l/A\\\rho =RA/l=5\times 10^{3}\times 3.14/50=314\Omega cm\\k=1/\rho =1/314\ Scm^{-1}$

The value of conductivity approximately is 1/300.

$Molar\ conductivity=k\times 1000/M=1000\times 1/314\times 0.08=39.808Scm^{2}mol^{-1}$