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Q.

Resistance of a conductivity cell (Cell constant 129 m–1) filled with 74.5 ppm solution of KCl is 100  Ω (labelled as solution 1). When the same cell is filled with KCl solution of 19 ppm, the resistance is 50  Ω (labelled as solution 2). The ratio of molar conductivity of solution 1 and solution 2 is i.e., 12=x×103. The value of x is (nearest integer)

Given, molar mass of KCl is 74.5  g  mol1

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Detailed Solution

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A=129   m1

KCl solution 1

74.5 ppm, R1=100  Ω, M1= 1x 10 - 3, cell constant=129

KCl solution 2

149 ppm, R2=50  Ω, M2= 2 x 10 - 3, cell constant=129

 

Here,ppm1ppm2=M1M2(W1/W2V×VW2/M0)

12=κ1×1000M1κ2×1000M2

=κ1×M2κ2×M1

=50100×2

=12=1,000×103

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