Home
Class 12
CHEMISTRY
^^(m) of 0.05 M weak electrolyte is "50 ...

`^^_(m)` of 0.05 M weak electrolyte is `"50 Sm"^(2)mol^(-1),0^(@)` of it is `"440 Sm"^(2)" mol"^(-1)`. Calculate `alpha` (degree of dissociation) of the electrolyte. 

Text Solution

Verified by Experts

Degree of dissociation of an electrolyte is given by
`alpha=(^^_(m))/(^^_(m)^(@))`
`"Given :  "^^_(m)="50 sm"^(2)" mol"^(-1)`
`^^_(m)^(@)="440 sm"^(2)" mol"^(-1)`
`"conc "=0.05m`
`alpha=(50)/(440)(cancel(sm^(2)))/(cancel(sm^(2)))(cancel(mol^(-1)))/(cancel(mol^(-1)))`
`alpha=0.
Promotional Banner

Topper's Solved these Questions

  • II PUC CHEMISTRY (P.U. BOARD LATEST MODEL QUESTION PAPER - 2)

    SUNSTAR PUBLICATION|Exercise PART - C|10 Videos

  • II PUC CHEMISTRY (P.U. BOARD LATEST MODEL QUESTION PAPER - 2)

    SUNSTAR PUBLICATION|Exercise PART - D|28 Videos

  • II PUC CHEMISTRY (P.U. BOARD LATEST MODEL QUESTION PAPER - 2)

    SUNSTAR PUBLICATION|Exercise PART - D|28 Videos

  • II PUC CHEMISTRY (P.U. BOARD LATEST MODEL QUESTION PAPER - 1)

    SUNSTAR PUBLICATION|Exercise PART - D|25 Videos

  • II PUC CHEMISTRY (P.U. BOARD LATEST MODEL QUESTION PAPER - 3)

    SUNSTAR PUBLICATION|Exercise PART - D|30 Videos

Similar Questions

Explore conceptually related problems

Equivalent conductivity of acetic acid at infinite dilution is 39.7 and for 0.1 M acetic acid the equivalent conductance is "5.2 mho.cm"^(2)."gm.equiv."^(-1) . Calculate degree of dissociation, H^(+) ion concentration and dissociation constant of the acid.

The K_c of an acid is 3.2 xx 10^(-5) . The degree of dissociation of the acid at concentration of 0.2 M is :

Resistance of 0.2 M solution of an electrolyte is 50Omega . The specific conductance of the solution is 1.4 S m^(-1) . If resistance of the 0.5 M solution of the same electrolyte is 280 Omega . The molar conductivity of 0.5 M solution of the electrolyte in Sm^(2)"mol"^(-1) is

Resistance of 0.2 M solution of an electrolyte is 50Omega . The specific conductance of the solution is 1.3 S m^(-1) . If resistance of the 0.4 M solution of the same electrolyte is 260 Omega , its molar conductivity is :

1 g of monobasic acid in 100 g of water lowers the freezing point by 0.168^(@) . If 0.2 g of same acid requires 15.1 mL mol^(-1) of N//10 alkali for complete neutralization, calculate the degree of dissociation of acid. K'_(f) for H_(2)O is 1.86 K mol^(-1) kg .

Heat of neutralisation of a weak acid HA by NaOH is -"12.13 kJ mol"^(-1) . Calculate the enthalpy of ionization of HA.

The molar conductance of a 0.1 M solution of an electrolyte was found to be 400 ohm^(-1) cm^(2) "mol"^(-1) . The cell constant of the cell is 0.1cm^(-1) . The resistance of the solution is :