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  • Resistance of a conductivity cell <img alt="\mathrm{ (cell \; constant \; 129 \mathrm{~m}^{-1} )}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%20%28cell%20%5C%3B%20constant%20

Resistance of a conductivity cell \mathrm{ (cell \; constant \; 129 \mathrm{~m}^{-1} )} filled with \mathrm{ 74.5\: \mathrm{ppm}} solution of \mathrm{ \mathrm{KCl}} is \mathrm{ 100\; \Omega} (labelled as solution 1). When the same cell is filled with \mathrm{ \mathrm{KCl}} solution of \mathrm{149\; \mathrm{ppm},} the resistance is \mathrm{ 50 \; \Omega} (labelled as solution 2). The ratio of molar conductivity of solution 1 and solution 2 is i.e.\mathrm{ \frac{\wedge_{1}}{\wedge_{2}}=x \times 10^{-3}. }The value of \mathrm{ x} is _______________. (Nearest integer)
Given, molar mass of \mathrm{ \mathrm{KCl}} is \mathrm{ 74.5 \mathrm{~g} \mathrm{~mol}^{-1}.}

Option: 1

1000


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\mathrm{Cell \: constant (G)= \frac{l}{A}= RK= 129\, m^{-1}}
\mathrm{(conductivity)K= \frac{G}{R}= \frac{129}{R}}
And we know \mathrm{\wedge _{m}= \frac{K\times 1000}{M}}

So, For 74.5 ppm  solution of \mathrm{KCl}
\mathrm{ppm_{1}= 74.5}
\mathrm{R_{1}= 180\Omega ,\wedge _{m1}= \frac{K_{1}\times 1000}{M_{1}}}

and For 149 ppm solution of \mathrm{KCl}
\mathrm{ppm_{2}= 149\, ppm}
\mathrm{R_{2}= 50\Omega }
\mathrm{\wedge _{m2} = \frac{K_{2}\times 1000}{M_{2}}}


 \mathrm{Now ,\; \frac{M_{1}}{M_{2}}= \frac{ppm_{1}}{ppm_{2}}= \frac{74.5}{145.0}}
\mathrm{and \; \frac{\wedge _{m1}}{\wedge _{m2}}= \frac{K_{1}}{K_{2}}= \frac{1}{M_{1}R_{1}}\times \frac{M_{2}R_{2}}{1}}

 
\mathrm{\Rightarrow \frac{\wedge _{m1}}{\wedge _{m2}}= \frac{R_{2}}{R_{1}}\times \frac{149.0}{74.5}= \frac{50}{100}\times 2= 1= 1000\times 10^{-3}}
Ans = 1000
 

Posted by

shivangi.shekhar

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