The equivalent conductance of 0.1 normal solution og MgCl2 is 97.1 ohm-1cm2eq-1(unit) at 25 degree C .A cell with electrodes that are 1.5cm square in surface area and 0.5cm apart ,0.1N MgCl2 sol.How much current will flow when potential difference between the electrodes is 5Volts..(final answer is I=0.1456A)
plz ans in a simple way..
In this question, we have the values of the potential difference and we have to calculate the value of the current. We are given that the equivalent conductance of 0.1 N MgCl2 solution is 97.1 (ohm)-1(equi)-1cm2. From here we will calculate the value of the conductivity.
Now we have the relation
λ = (κ / c)
where λ is the equivalent conductivity, κ is the conductivity and c is the concentration of the solution. Substituting the values of λ and c in the above equation, we get
κ = λ X c
= 97.1 equi)-1cm2 X 0.1 X 10-3 equivalents cm-3
[normality = number of gram equivalents of solute / Volume of solution in litres]
= (number of gram equivalents of solute X 10-3) / Volume of solution in cm-3 ]
This gives κ = 97.1 X 10-4(ohm)-1X cm-1
Now κ = G (l/A)
where G is the conductance of the solution, l is the distance between the electrodes of the cell and A is the area of the electrode plates. Substituting the value of κ = 97.1 X 10-4(ohm)-1X cm-1, A = 1.5 cm2 and l = 0.5 cm, we get
G = 97.1 X 10-4 X (1.5 / 0.5)
= 291.3 X 10-4
= 2.913 X 10-2 (ohm)-1
Now G is equal to 1 / R, where R is the resistance. From ohm's law we have
V = I X R
where V is the potential difference, R is the resistance and I is the current. Thus
I = V X ( 1 / R)
= V X G
= 5 X 2.913 X 10-2
= 0.1456 A