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A mixture of KBr and NaBr weighing 0.560...

A mixture of KBr and NaBr weighing 0.560 g was treated with aqueous `Ag^(+)` and all the bromide ion was recovered as 0.970 g of pure AgBr . What was the fraction by weight of KBr in the sample ?
`(K =n 39, Br = 80 , Ag = 108 , Na = 23 )`

Text Solution

Verified by Experts

`{:("KBr + NaBr + " AG^(+) to " " AGBr),(" xg (0.56-x) 0.97 g"):}`
Since Br atoms are conserved, applying POAC for Br atoms,
Moles of Br in KBr + moles of Br in NaBr = moles of Br in AgBr
or ` 1 xx ` moles of KBr +` 1 xx ` moles of NaBr = `1 xx ` moles of AGBr
x = 0.1332 g
Fraction of KBr in the sample `= (01332)/(0.560) = 0.2378`
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