Molar conductances of $B a C l_{2}, H_{2} S O_{4}$ and HCl at infinite dilutions are $x_{1}, x_{2}$ and $x_{3}$ respectively. Equivalent conductance of $B a S O_{4}$ at infinite dilution will be :

Q. The equivalent conductivity of solutions of

B a C l_{2}, H_{2} S O_{4} and H C l

, are

x_{1}, x_{2} and x_{3} {Scm}^{2} {eq}^{- 1}

, respectively at infinite dilution. If the conductivity of saturated

B a S O_{4}

solution is

{x Scm}^{- 1}

, then the

K_{s p}

B a S O_{4}

is:

Q. Molar conductances of

B a {C l}_{2}, H_{2} {S O}_{4}

and

H C l

are

x_{1}, x_{2}

and

x_{3}

S {c m}^{2}

{m o l}^{- 1}

at infinite dilution. If specific conductance of saturated

B a {Z O}_{4}

solution is of

y

S {c m}^{- 1}

, then

k_{s p}

B a {S O}_{4}

is:

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Equivalent conductance of BaCl2,H2SO4 and HCl are x1,x2 and x3Scm2equiv−1 at infinite dilution.If specific conductance of saturated BaSO4 solution is of yScm1 then Ksp of BaSO4 is:

The correct option is A 103y2(x1+x2−2x3)Λeq(BaSO4)=Λeq(BaCl2)+Λeq(H2SO4)−2Λeq(HCl) =(x1+x2−2x3)Scm2eq−1∵Λeq=K×1000N⇒N=(y×103x1+x2−2x3)(∵nf=2)∴M={y×1032(x1+x2−2x3)}∴Ksp=(M)2={y2×1064(x1+x2−2x3)2}

Equivalent conductance of $B a C l_{2}, H_{2} S O_{4}$ and $H C l$ are $x_{1}, x_{2}$ and $x_{3} S c m^{2} e q u i v^{- 1}$ at infinite dilution.If specific conductance of saturated $B a S O_{4}$ solution is of $y S c m^{1}$ then $K_{s p}$ of $B a S O_{4}$ is: