A mixture of `Xe` and `F_(2)` was heated and the white solid so formed reacted with `H_(2)` to give `81 mL of Xe` at `STP` and `HF`. The `HF` formed required `68.43 mL` of `0.3172 M NaOH` for complete neutralisation. Determine empiriacal formula of white solid.

Explanation

`2Xe+nF_(2)rarr2XeF_(n)`
`2XeF_(n)+nH_(2)rarr2Xe+2nHF`
Mole of `Xe` formed `=(81)/(22400)=3.6xx10^(-3)`
Meq. Of `HIF` formed `=Meq."of" NaOH`
`=68.43xx0.3172`
`:. m` of `HF` formed `=68.43xx0.3172`
Mole of `HF` formed`=(68.43xx0.3172)/(1000)`
`=21.7xx10^(-3)`
`("Mole of"Xe)/("Mole of"HF)=(2)/(2n)`
`:. (3.6xx10^(-3))/(21.7xx10^(-3))=(2)/(2n)`
`:. n=6`
Thus, compound is `XeF_(6)`.