Excess `F_(2)(g)` reacts at `150^(@)C` and 1.0 atm pressure with `Br_(2)(g)` to give a compound `BrF_(n)`. If 423 mL of `Br_(2)(g)` at the same temperature and pressure produced 4.2 g of `BrF_(n)`, what is n? [Atomic mass Br =80, F =19 ]
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The correct Answer is:
5
`n_(Br_(2))=(1xx0.423)/(0.0821xx423)=(1)/(82.1) "mole"` POAC on 'Br' `(1)/(82.1)xx2=(4.2)/(80+19n)xx1` `n=5`
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