0.1 M solution of AgNO ₃ is taken in a Conductivity cell and a potential difference of 40 ~V is applied across the ends of a column of this solution whose diameter is 4.0 ~cm and length of the column is 12 ~cm . The current used is 0.4 ~A . The Molar conductivity of the solution is _________.

Step-by-step solution

The resistance R of the solution is given by Ohm's law: R = V I = 40 V 0.4 A = 100 . The area of cross-section A of the column is r^2 . Given diameter d = 4.0 cm , the radius r = 2.0 cm . Thus, A = (2.0 cm )^2 = 4 cm ^2 12.566 cm ^2 . The cell constant G^ * is given by l A , where l = 12 cm . So, G^ * = 12 4 = 3 cm ⁻¹ 0.9549 cm ⁻¹ . The conductivity is given by = G^ * R = 3/ 100 = 3 100 S cm ⁻¹ 0.009549 S cm ⁻¹ . The molar conductivity _m is calculated using the formula _m = 1000 C , where C = 0.1 M . Substituting