COMEDK2021ChemistryElectrochemistry
The specific conductivity of a solution containing 1.0 ~g of anhydrous BaCl ₂ in 200 ~cm ³ of the solution has been found to be 0.0058 Scm ⁻¹ . The molar and equivalent conductivity of the solution respectively are
Options
- A120.83 ~S ~cm ² eq ⁻¹ and 241.67 ~S ~cm ² eq ⁻¹
- B150.5 ~S ~cm ² eq ⁻¹ and 289.7 ~S ~cm ² eq ⁻¹
- C241.6 ~S ~cm ² ~mol ⁻¹ and 120.83 ~S ~cm ² eq ⁻¹
- D248.6 ~S ~cm ² ~mol ⁻¹ and 180.3 ~S ~cm ² eq ⁻¹
Correct answer
C. 241.6 ~S ~cm ² ~mol ⁻¹ and 120.83 ~S ~cm ² eq ⁻¹
Step-by-step solution
The molar mass of BaCl ₂ is 137.3 + 2 35.5 = 208.3 ~g/mol . The number of moles of BaCl ₂ is n = 1.0 ~g 208.3 ~g/mol 0.0048008 ~mol . The molarity M of the solution is M = n V( in L ) = 0.0048008 ~mol 0.200 ~L = 0.024004 ~M . The molar conductivity _m is calculated as _m = 1000 M = 0.0058 ~S cm ⁻¹ 1000 ~cm ^3 L ⁻¹ 0.024004 ~mol L ⁻¹ 241.626 ~S cm ^2 mol ⁻¹ . For BaCl ₂ , the n-factor is 2 (since BaCl ₂ Ba ²⁺ + 2 Cl ⁻ ). The equivalent conductivity _ eq is calculated as _ eq = _m n -factor = 241.626 2 120.813 ~S cm