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The conductivity of a saturated solutior...

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

A

`2 xx 10^(-10)M^(3)`

B

`2 xx 10^(-12)M`

C

`8 xx 10^(-12)M`

D

`8 xx 10^(-14)M`

Text Solution

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The correct Answer is:
C
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The conductivity of a solution of AgCl at 298 K is found to be 1.382xx10^(-6)Omega^(-1)cm^(-1) the ionic conductance of Ag^(+) and Cl^(-) at infinite dilution are 61.9Omega^(-1)cm^(2)col^(-1) ad 76.3Omega^(-1)cm^(2)mol^(-1) respectively the solubility of AgCl is

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The specific conducitvity of a saturated solution of AgCI is 3.40 xx 10^6 "ohm"^(-1) cm^(-1) at 25^@C . If lambda_(Ag+) = 62.3 "ohm"^(-1) cm^2 "mol"^(-1) and lambda_(CI-) = 67.7 "ohm"^(-1) cm^2 "mol"^(-1) , the solubility of AgCI at 25^@C is.

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) )

Calculate Lambda_(m)^(oo) for acetic acid, given, Lambda_(m)^(oo)(HCl)=426 Omega^(-1)cm^(2)mol^(-1),Lambda_(m)^(oo)(NaCl)=126Omega^(-1)cm^(2)mol^(-1) , Lambda_(m)^(oo)(CHCOONa)=91Omega^(-1)cm^(2)mol^(-1)

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