Question

0.5N Al2 (SO4)3 solution is present between two
platinum electrode
that
are 2cm x 4cm. the
resistance
of solution 200 ohm. Calculate molar
conductance of
solution.

​

Answer 1

Explanation:

The resistance of 0.01 N solution at `25^(@)C` is 200 ohm. Cell constant of the conductivity cell is unity. Calculate the equivalent conductance of the solution.

Hope it helps ya!!

Answer 2

Given info : 0.5N Al2 (SO4)3 solution is present between two platinum electrode that are 2cm x 4cm. the resistance of solution 200 ohm.

To find : the molar conductance of solution is ..

solution : specific conductance, k = conductance × cell constant

here resistance, R = 200 ohm

so conductance, C = 1/resistance = 1/200 mho

cell constant = l/A = distance between two electrodes/area of each electrode

here distance between electrodes hasn't given. question should mention this one.

let distance between electrodes is 1 cm.

then, cell constant = 1cm/(2cm × 4cm) = 100/8 m¯¹

= 12.5 m¯¹

now specific conductance, k = 1/200 × 12.5

= 6.25 × 10¯² mho m¯¹

we know, n factor of Al2(SO4)3 = 6

so molarity × n - factor = normality

⇒molarity, M = 0.5/6 M

now molar conductivity = k × 1000/M

= (6.25 × 10¯² × 1000)/(0.5/6)

= 62.5 × 6/0.5

= 125 × 6

= 750 Seimen m² mol