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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. |
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Answer» `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)`. |
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