Home
Class 12
CHEMISTRY
If conductivity of water used to make sa...

If conductivity of water used to make saturated of AgCl is found to be `3.1xx10^(-5)Omega^(-1)`
`cm^(-1)` and conductance of the solution of `AgCl=4.5xx10^(-5)Omega^(-1)cm^(-1)`
if `lamda_(M)^(0)AgNO_(3)=200Omega^(-1)cm^(2)` `"mole"^(-1)`
`lamda_(M)^(0)NaNO_(3)=310Omega^(-1)cm^(2)` `"mole"^(-1)`
calculate `K_(SP)` of `AgCl`

Promotional Banner

Similar Questions

Explore conceptually related problems

The conductivity of saturated solution of AgCl is found to be 1.86 xx 10^(-6) ohm^(-1)cm^(-1) and that of water is 6xx 10^(-8) ohm^(-1)cm^(-1) . If lamda^0 AgCl is

At 298 K, the conductivity of a saturated solutioni of Agcl in water is 2.6xx10^(-6)ohm^(-1)cm^(-1) givem lamda_(m)^(oo)(Ag^(+))=63ohm^(-1)cm^(2)mol^(-1) and lamda_(m)^(oo)(Cl^(-))=67ohm^(-1)cm^(2)mol^(-1) The solubility product of AgCl is

The specific conductance of saturated solution of AgCl is found to be 1.86xx10^(-6) ohm^(-1) cm^(-1) and that of water is 6 xx 10^(-8) ohm^(-1) cm^(-1) . The solubility of agCl is ... . Given, Lambda_(AgCl)^(@)=137.2 ohm^(-1) cm^(2) eq^(-1)

Calculate lamda_(m)^(0) of oxalic acid, given that lamda_(eq)^(0)Na_(2)C_(2)O_(4)=400Omega^(-1)cm^(2)aq^(-1) lamda_(m)^(0)H_(2)SO_(4)=700Omega^(-1)cm^(2)"mole"^(-1) lamda_(eq)^(0)Na_(2)SO_(4)=450Omega^(-1)cm^(2)eq^(-1)

Calculate lamda_(m)^(0) of oxalic acid, given that lamda_(eq)^(0)Na_(2)C_(2)O_(4)=400Omega^(-1)cm^(2)aq^(-1) lamda_(m)^(0)H_(2)SO_(4)=700Omega^(-1)cm^(2)"mole"^(-1) lamda_(eq)^(0)Na_(2)SO_(4)=450Omega^(-1)cm^(2)eq^(-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 solibility of AgCl at 25^(@)C is:

The resistance of 0.1 N solution of a salt is found to be 2.5xx10^(3) Omega . The equivalent conductance of the solution is (Cell constant =1.15 cm^(-1) )

The resistance of 0.1 N solution of a salt is found to be 2.5xx10^(3) Omega . The equivalent conductance of the solution is (Cell constant =1.15 cm^(-1) )

The resistance of 0.1 N solution of salt is found to be 2.5 xx 10^(3) Omega . The equivalent conductance of solution (cell constant = 1.15 cm^(-1) ) in Omega^(-1) cm^(2) eq^(-1) is