Question

The resistance of 0.01 n solution of an electrolyte was found to be 210 ohm at 298 k, using a conductivity cell of cell constant 0.66 cm–1. the conductivity of solution is

Answer 1

The resistance of 0.01 N solution of an electrolyte AB at 328 K is 100 ohm. The specific conductance of solution is (cell constant = 1cm-1)

Answer 2

Given:

Resistance (R) = 210 ohm

Cell constant (K) = 0.66 $cm^{-1}$

Concentration = 0.01 N

To find:

Conductivity of solution = ?

Formula to be used:

λ = $\frac{100\kappa}{C}$

$\kappa = \frac{K}{R}$

Calculation:

$\kappa = \frac{K}{R}$

$\kappa = \frac{0.66 cm^{-1}}{210 ohm}$

$\kappa = 3.14\times 10^{-3} ohm^{-1}cm^{-1}$

Substituting the $\kappa$ value in the formula it gives:

λ = $\frac{100\kappa}{C}$

λ = $\frac{100\times3.14\times 10^{-3} ohm^{-1}cm^{-1}}{0.01 N}$

λ = 314.28$ohm cm^{2} eq^{-1}$

Conclusion:

The conductivity of solution is calculated as 314.28$ohm cm^{2} eq^{-1}$.

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