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At infinite dilution the molar conductan...

At infinite dilution the molar conductance of `Al^(+3)` and `SO_(4)^(-2)` ion are 189 and `160 Omega^(-1) cm^(2) "mole"^(-1)` respectively. Calculate the equivalent and molar conductivity at infinite dilute of `Al_(2)(SO_(4))_(3)`.

A

150

B

180

C

190

D

270

Text Solution

Verified by Experts

The correct Answer is:
D
`{:(AgA(s)hArr,Ag^(+)+,A^(-)),(,S_(1)+S_(2),S_(1)):}`
`((K_(sp))_(AgA))/((K_(sp))_(AgB))=(S_(1))/(S_(2))=(3)/(1)`
`{:(AgB(s)hArr,Ag^(+)+,B^(-)),(,S_(1)+S_(2),S_(2)):}`
`S_(1)+S_(2)= sqrt((K_(sp))_(AgA)+(K_(sp))_(AgB))=2xx10^(-7)`
On solving `S_(1)=(3)/(2)xx10^(-7)` and `S_(2)=(1)/(2)xx10^(-7)`
`K_("solution")=k_(Ag^(+))+k_(A^(-))+k_(B^(-))`
`375xx10^(-4)=60xx2xx10^(-10)+(3)/(2)xx10^(-10)xx80+(1)/(2)xx10^(-10)xxlambda_(B^(-))^(oo)`
`375=120+120+(lambda_(B^(-))^(oo))/(2)`
`lambda_(B^(-))^(oo)=270` Scm `mol^(-1)` .
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