At 298K the conductivity of a saturated solution of AgCl in water is `2.6times10^(-6)Scm^(-1)` .Its solubility product at 298K (given: `lambda^(oo)(Ag^(+))`=63.0`Scm^(2)mol^(-1)`` lambda^(oo)(Cl^(-))`=67.0`Scm^(2)mol^(-1) `

The specific conductivity of a saturated solution of AgCl is 3.40xx10^(-6) ohm^(-1) cm^(-1) at 25^(@)C . If lambda_(Ag^(+)=62.3 ohm^(-1) cm^(2) "mol"^(-1) and lambda_(Cl^(-))=67.7 ohm^(-1) cm^(2) "mol"^(-1) , the solubility of AgC at 25^(@)C is:

The conductivity of 0.20M solution of KCl at 298K is 0.0248 S cm^(-1) . Calculate its molar conductivity.

We have taken a saturated solution of AgBr , whose K_(sp) is 12xx10^(-14). If 10^(-7)M of AgNO_(3) are added to 1L of this solutino, find the conductivity ( specific conductance ) of the solution in terms of 10^(-7)S m^(-1) units. Given : lambda^(@)._((Ag^(o+)))=6xx10^(-3)S m^(2) mol^(-1) lambda^(@)._((Br^(c-)))=8xx10^(-3)S m^(2)mol^(-1) lambda^(@)._((NO_(3)^(C-)))=7XX10^(-3)S m^(2) mol^(-1)

Conductivity of a saturated solution of a sparingly soluble salt AB at 298K is 1.85× 10^(−5) Sm^(−1) . Solubility product of the salt AB at 298K is : [Given Λ^(o)m (AB)=140× 10 ^(−4)Sm^2mol^(−1) ]

The conductivity of a saturated solution of a sparingly soluble salt MX_2 is found to be 4 xx 10^(-5) Omega^(-1) cm^(-1)." If "lambda_(m)^(oo) ( M^(2+))=50 Omega^(-1) cm^(2) mol^(-1) and lambda^(oo) (X^(-))= 50 Omega^(-1) cm^(2) mol^(-1) , the solubility product of the salt is about

The molar conductivity of 0.25 mol L^(-1) methanoic acid is 46.1 S cm^(2) mol^(-1) . Calculate the degree of dissociation constant. Given : lambda_((H^(o+)))^(@)=349.6S cm^(2)mol^(-1) and lambda_(HCOO^(-))^(@)=54.6Scm^(2)mol^(-1)

If conductivity of 0.1 mol/dm^3 solution of KCl is 1.3 * 10^(-2) S cm^(-1) at 298 K then its molar conductivity will be

We have taken a saturated solultion of AgBr, K_(SP) is 12 xx 10^(-4) . If 10^(-7) M of AgNO_(3) are added to 1 L of this solution, find conductivity (specific conductance) of this solution in term of 10^(-7) Sm^(-1) units. Given, lambda_((Ag^(+)))^(@) = 6 xx 10^(-3) Sm^(2) mol^(-1) , lambda_((Br^(-)))^(@) = 8 xx 10^(-3) Sm^(2) mol^(-1) , lambda_((NO_(3)^(-)))^@ = 7 xx 10^(-3) Sm^(2) mol^(-1) . Neglect the contribution of solvent.

Conductivity of an aqueous solution of 0.1M HX (a weak mono-protic acid) is 5 xx 10^(-4) Sm^(-1) Given: ^^_(m)^(oo) [H^(+)] = 0.04 Sm^(2) mol^(-1): ^^_(m)^(oo) [X^(-)] = 0.01 Sm^(2)mol^(-1) Find pK_(a)[HX]

Molar conductivity of aqueous solution of HA is 200Scm^(2)mol^(-1),pH of this solution is 4 Calculate the value of pK_(a)(HA) at 25^(@)C . Given ^^_(M)^(oo)(NaA)=100scm^(2)mol^(-1), ^^_(M)^(oo)(HCl)=425Scm^(2)mol^(-1), ^^_(M)^(oo)(NaCl)=125 Scm^(2)mol^(-1)