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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:-A. `4.32 xx 10^(-18)`B. `1.8 xx 10^(-9)`C. `8.64 xx 10^(-13)`D. None of these |
Answer» Correct Answer - A `lambda_(eq)+K +(1000)/(N)` `1.50xx10^(-4)xx10^(4)=9xx10^(-5)xx10^(-2)xx(1000)/(N)` `N= 6xx10^(-5)` `S=M=(N)/(n_(f))=(6xx10^(-5))/(3)=2xx10^(-5)mol//L` . `{:(Ag_(3)PO_(4)=,3 Ag^(+)+,PO_(4)),(,3S,S):}` `K_(sp)=(3S)^(3).S=27.S^(4)=27xx(2xx10^(-5))^(4)` `=4.32xx10^(-18)` . |
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