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The specific conductance of a N/10 kCl at  25^{\circ} \mathrm{C}  is  0.0248 \mathrm{~ohm}^{-1} \mathrm{~cm}^{-1}  The resistance of cell containing solution at same temp was as found be 50 ohms. The cell constant will be 

Option: 1

1. 24 


Option: 2

1.87 


Option: 3

2.84
 


Option: 4

 2.24


Answers (1)

best_answer

\mathrm{ K=\frac{1}{R}\left(\frac{l}{a}\right) }
\mathrm{ 0.0248=\frac{1}{50}\left(\frac{l}{a}\right) }

\mathrm{ \left(\frac{l}{a}\right)=0.0248 \times 50 }


\mathrm{ \frac{l}{a}=1.24 }

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Divya Prakash Singh

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