Answer
$KI$ $0.1M$ has a higher concentration of $I^-$.
Work Step by Step
$KI$ $0.1M$:
$[I^-] = 1 \times [KI]$
$[I^-] = 0.1M$
$ZnI_2$ $0.040M$:
$[I^-] = 2 \times [ZnI_2]$
$[I^-] = 0.080M$
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