Home
Class 12
CHEMISTRY
At 298 K, the specific conductance of 0....

At 298 K, the specific conductance of 0.1 M acetic acid solution was found to be `0.00163Omega^(-1)cm^(-1)`. Calculate the degree of dissociation and dissociation constant of the acid if its molar conductance at infinite dilution is `390.7 Omega^(-1) cm^(2)mol^(-1)`.

Promotional Banner

Similar Questions

Explore conceptually related problems

The specific conductivity of 0.1 M solution of a weak acid at 298 K is 0.00159 "ohm^(-1) cm^(-1) . Calculate the degree of dissociation of the acid if its molar conductance at infinite dilution is 350.5 "ohm"^(-1) cm^(2) mol^(-1) .

The conductivity of 0.001 M acetic is 4 xx 10^(-5)S//cm. Calculate the dissociation constant of acetic acid, if molar conductivity at infinite dilution for acetic is 390 S cm^(2)//mol.

The conductivity of 0.001 M acetic is 4 xx 10^(-5)S//cm. Calculate the dissociation constant of acetic acid, if molar conductivity at infinite dilution for acetic is 390 S cm^(2)//mol.

The conductivity of 0.001M acetic acid is 4 times 10^-8 S//cm . Calculate the dissociation constant of acetic acid, if molar conductivity at infinite dilution for acetic acid is 390 S cm^2//mol .

The conductivity of 0.001 M actic acid is 4 xx 10^(-5) S //cm . Calculate the dissociation constant acetic acid, if molar conductivity at infinite dilution for acetic acid is 390 S cm^(2) // mol.

Kohlrausch’s law Is useful for determining lambda_m^0 of weak electrolytes.The molar conductance of 0.3 M acetic acid Is 176.6 ohm^(-1) cm^2 mol^-1 . Calculate its degree of dissociation if molar conductance at infinite dilution is 384 ohm^-1 cm^2 mol^-1 .

The equivalent conductance of 0.1 N acetic acid is 5 cm^2 ohm^-1 gm eq^-1 and at infinite dilution is 390 cm^2ohm^-1gm eq^-1 . Calculate the degree of dissociation of acetic acid.