Question

conductivity of a solution containing 1 gram of anhydrous bacl2 in centimetre cube what has been found to be 0.0058 7 centimetre inverse what are the molar conductance and equivalent conductance of the solution (if weight of Ba =137 and cl=35.5​

Answer 1

Answer:

Explanation:

Mass of Bacl2 = 1 gram  (Given)

Let the  mass of solution = 200mL

Conductivity of the solution (k) = 0.00587 cm (Given)

Molarity = Moles of salute/ Volume of solution

Molar mass of BaCl2 = 137 + 2 ×35.5 = 208 g mol-1

Number of moles of BaCl2 in 200 cm3 of solution = 1/208

Volume of solution that contains 1 mol of BaCl2 (V) = 200 × 208cm³

Molar conductivity = Am × kV

= Am = 0.0058 × 200 × 208

= 241.28 cm² mol-1

Therefore, the molar conductance of the element is 241.28 cm² mol-1.