The equivalent conductance of 200ml HCL soln has found to be 2×10^3 times its specific conductance. How much water should be added to the soln so that eq conductance becomes 5×10^3 its specific conductance?

Question

Vikas TU · Answer

Eq. Conductance = 2000K = K*1000/N => N1 = 0.5 For 5000K eq. conductance, 5000K = 1000K/N N2 = 1/5 Now, N1V1 = N2V2 0.5*200 = 0.2*V2 V2 = 100/0.2 => 500 ml.

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The equivalent conductance of 200ml HCL soln has found to be 2×10^3 times its specific conductance. How much water should be added to the soln so that eq conductance becomes 5×10^3 its specific conductance?

The equivalent conductance of 200ml HCL soln has found to be 2×10^3 times its specific conductance. How much water should be added to the soln so that eq conductance becomes 5×10^3 its specific conductance?

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The equivalent conductance of 200ml HCL soln has found to be 2×10^3 times its specific conductance. How much water should be added to the soln so that eq conductance becomes 5×10^3 its specific conductance?

The equivalent conductance of 200ml HCL soln has found to be 2×10^3 times its specific conductance. How much water should be added to the soln so that eq conductance becomes 5×10^3 its specific conductance?

1 Answers

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