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Resistance of a conductivity cell (cell constant \mathrm{129 m^{-1}} ) filled with 30.6ppm solution of cell is \mathrm{110 \Omega} {laballed as solution 1} When the same cell is filled with Kl sol. of 168ppm, the resistance \mathrm{20 \Omega} {labelled as sol 2} The Calculate the ratio of molar conductivity of sol 1 and sol 2.

Option: 1

1.684


Option: 2

1.497


Option: 3

1.968


Option: 4

1.369


Answers (1)

best_answer

KCl sol 1:-
\mathrm{ {30.6 ppm } R_1=110 \Omega }
KCl sol 2:-
\mathrm{ 168 \mathrm{ppm} R_2=30 \Omega }


Here;

\mathrm{\frac{p p m_1}{p p m_2}=\frac{M_1}{M_2} }
\mathrm{K=\frac{1}{D} G^* ~here ~G^* } is constant

\mathrm{\frac{\Lambda_1}{\Lambda_2} =K_1 \times \frac{1000}{M_2}}

            \mathrm{K_2 \times \frac{1000}{M_2}}

\mathrm{\frac{K_1}{K_2} \times \frac{M_2}{M_1}=\frac{R_2}{R_1} \times \frac{M_2}{M_1}}

\mathrm{\frac{30}{110} \times \frac{168}{30.6}}

\mathrm{\frac{\Lambda_1}{\Lambda_2}=1.497}

Posted by

Suraj Bhandari

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