At infinite dilution, the percentage ion...

At infinite dilution, the percentage ionisation of both strong and weak electrolytes is

`1%`

`20%`

`50%`

`100%`

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The correct Answer is:

According to Ostwald's dilution law because degree of ionization is directly proportional to the dilution.

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At infinite dilution, the percentage dissociation of both weak acid and weak base is:

Differentiate strong and weak electrolytes

At 0.04 M concentration, the molar conductivity of solution of an electrolyte is 5000 Omega^(-1)cm^(2) mol^(-1) while at 0.01 M concentration the value is 5100 Omega^(-1)cm^(2)mol^(-1) . Making necessary assumption (Taking it as strong electrolyte) find the molar conductivity at infinite dilution and write percentage dissociation of strong electrolyte at 0.04 M.

The equivalent conductivity of 0.1 M weak acid is 100 times less than that at infinite dilution. The degree of dissociation of weak electrolyte at 0.1 M is.

At infinite dilution, when the dissociation of electrolyte is complete, each ion makes a definite contribution towards the molar conductance of electrolyte, irrespective of the nature of the other ion with which it is associated. the molar conductance of an electrolyte at infinite dilution can be expressed as the sum of the contributions from its individual ions. A_(x)B_(y) rarr xA^(y+)+yB^(x-) Lambda_(m)^(@)(A_(x)B_(y))=xlambda_(A^(y+))^(@)+ylambda_(B^(x-))^(@) where, x and y are the number of cations and anions respectively. The degree of ionisation 'alpha' of weak electrolyte can be calculated as : alpha=Lambda_(m)/Lambda_(m)^(@) The molar conductances at infinite dilution for electrolytes BA and CA are 140 and 120 ohm^(-1) cm^(2) mol^(-1) . If the molar conductance at infinite dilute dilution of BX is 198 ohm^(-1) cm^(2) mol^(-1) , then at infinite dilution, the molar conductance of CX is :

At infinite dilution, when the dissociation of electrolyte is complete, each ion makes a definite contribution towards the molar conductance of electrolyte, irrespective of the nature of the other ion with which it is associated. the molar conductance of an electrolyte at infinite dilution can be expressed as the sum of the contributions from its individual ions. A_(x)B_(y) rarr xA^(y+)+yB^(x-) Lambda_(m)^(@)(A_(x)B_(y))=xlambda_(A^(y+))^(@)+ylambda_(B^(x-))^(@) where, x and y are the number of cations and anions respectively. The degree of ionisation 'alpha' of weak electrolyte can be calculated as : alpha=Lambda_(m)/Lambda_(m)^(@) The unit of molar conductance of an electrolyte solution will be :

At infinite dilution, when the dissociation of electrolyte is complete, each ion makes a definite contribution towards the molar conductance of electrolyte, irrespective of the nature of the other ion with which it is associated. the molar conductance of an electrolyte at infinite dilution can be expressed as the sum of the contributions from its individual ions. A_(x)B_(y) rarr xA^(y+)+yB^(x-) Lambda_(m)^(@)(A_(x)B_(y))=xlambda_(A^(y+))^(@)+ylambda_(B^(x-))^(@) where, x and y are the number of cations and anions respectively. The degree of ionisation 'alpha' of weak electrolyte can be calculated as : alpha=Lambda_(m)/Lambda_(m)^(@) The ionic conductances of Al^(3+) and SO_(4)^(2-) ions at infinite dilution are x and y ohm^(-1) cm^(2) mol^(-1) respectively. If Kohlrausch's law is valid then molar conductance of aluminium sulphate at infinite dilution will be :

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