Assume 100% ionisation of the following in aq. solution of
(I) $[P t {(N H_{3})}_{6}] C l_{4}$ (II) $[C r {(N H_{3})}_{6}] C l_{3}$ (III) $[C o {(N H_{3})}_{4}] C l_{2}$ (IV) $NaCl$
Increasing order of conductivity is:

Question

Assume 100% ionisation of the following in aq. solution of (I) P t N H 3 6 C l 4 (II) C r N H 3 6 C l 3 (III) C o N H 3 4 C l 2 (IV) NaCl Increasing order of conductivity is:

TestHub · Accepted Answer

(I) P t N H 3 6 C l 4 :no of ions = 5 P t N H 3 6 C l 4 → P t N H 3 6 + 4 + 4 C l - (II) C r N H 3 6 C l 3 :no of ions = 4 C r N H 3 6 C l 3 → C r N H 3 6 3 + + 3 C l - (III) C o N H 3 4 C l 2 :no of ions = 3 C o N H 3 4 C l 2 → C o N H 3 4 2 + + 2 C l - (IV) NaCl: no of ions = 2 N a C l → N a + + C l - So Increasing order of conductivity would be: IV < III < II < I

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