1.

A mixture of KBr and NaBr weighing 0.560 g wastreated with aqueousAg^(+) and all the bromide ion was recoveredas 0.970 g ofpure AgBr . What was the fraction by weight of KBrin the sample ? (K =n 39, Br = 80 , Ag = 108 , Na = 23 )

Answer»

SOLUTION :`{:("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 = molesof 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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