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1.11g CaCl(2) is added to water forming ...

`1.11g CaCl_(2)` is added to water forming `500ml` of solution `20ml` of this solution is taken and diluted `10` folds Find moles of `Cl` ions in `2ml` of diluted solution .

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`(1.11)/(111) = 0.01 "mol" CaCl_(2)`
Moles of `CaCl_(2)` in 20ml solution `= (0.01)/(500) xx 20 = (0.01)/(25)`
In `200ml` solution moles of `CaCl_(2) = (0.01)/(25)` [Note Dilution does not change moles of solute]
In `2ml` of dilute solution moles of `CaCl_(2)=(0.01)/(25/(2000))xx2=(0.01)/(2500)=8xx10^(-6)` .
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