[ [10^(-6)" mole of "AgNO_(3)" are added in "1" Litre saturated solution of "AgX(K_(SP)=a times10^(-6)M^(2))" .Conductivity "],[" of this solution is "27times10^(-6)Sm^(-1),lambda_(Ag^(+))^(0)=6times10^(-3)Sm^(2)mol^(-1),lambda_(X^(-))^(0)=8times10^(-3)Sm^(2)mol^(-1)" ,"],[lambda_(NO_(3)^(-))^(0)=7times10^(-3)Sm^(2)mol^(-1)" .Determine value of a."]

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.

We have taken a saturated solution of AgBr.K_(sp) of AgBr is 12xx10^(-14) . If 10^(-7) mole of AgNO_(3) are added to 1 mitre of this solutio then the conductivity of this solution in terms of 10^(-7)Sm^(-1) units will be [given lamda((Ag^(+)))^(@)=4xx10^(-3)Sm^(2)mol^(-1)lamda_((Br^(-)))^(@)=6xx10^(-3)Sm^(2)mol^(-1). lamda_((NO_(3)^(-)))^(@)=5xx10^(-3)Smmol^(-1)) (A). 39 (B). 55 (C). 15 (D). 41

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)

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 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)

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]

The conductivity of a saturated solution of Ag_(3)PO_(4) is 9 xx 10^(-6) S m^(-1) and its equivalent conductivity is 1.50 xx 10^(-4) Sm^(2) "equivalent"^(-1) . Th K_(sp) of Ag_(3)PO_(4) is:-

If the rate constant k of a reaction is 1.6 times 10^–3(mol//l)(min^–1) . The order of the reaction is:

The conductivity of a 0.05 M solution of a weak monobasic acid is 10^(-3) S cm^(-1) , If lambda_(m)^(oo) for weak acid 500 Scm^(2) "mol"^(-1) , calculate Ka of weak monobasic acid :

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) ]