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  • If   (of sol )<img alt="\mathrm{=1192 \mathrm{scm}^2mol^{-

If  \mathrm{\Lambda ^{0}m} (of sol )\mathrm{=1192 \mathrm{scm}^2mol^{-1}}  and conductivity of it is \mathrm{3 \times 10^{-4} \mathrm{~cm}^{-1}} with concentration 0.001 , What is ionization constant ( Ka) of sol.

Option: 1

0.25


Option: 2

0.31


Option: 3

1.84


Option: 4

2.97


Answers (1)

best_answer

\mathrm{ \lambda_m=1000 \times \frac{K}{M} \\ }

\mathrm{ \lambda m=1000 \times \frac{3 \times 10^{-4}}{0.001}= 300 \mathrm{scm}^2 mol^{-1} }.

\mathrm{ x=\frac{\Lambda m}{\Lambda m^{\circ}}=\frac{300}{1192}=0.25 }

Posted by

jitender.kumar

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