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60ml of mixture of equal volumes of Cl(2...

`60ml` of mixture of equal volumes of `Cl_(2)` and an oxide of chloine, i.e., `Cl_(2)O_(n)` was heated and then cooled back to the original temperature. The resulting gas mixture was found to have volume of `75ml`. On treatment with `KOH` solution, the volume contracted to `15ml`. Assume that all measurements are made at the same temperature and pressure. Deduce the value of `n` in `Cl_(2)O_(n)`. The oxide of `Cl_(2)` n heating decomposes quantiatively to `O_(2)` and `Cl_(2)`.

Text Solution

Verified by Experts

Let the oxide is `Cl_(x).O_(y)`.
`Cl_(x)O_(y)to(x)/(2)Cl_(2)+(y)/(2)O_(2)`
Volume of `Cl_(x)O_(y)=` volume of `Cl_(2)=3 mL`
After cooling, volume `=75mL`
This corresponds to volume of `Cl_(2)` initially plus volume of `Cl_(2)` produced and `O_(2)` produced.
`thereforeV_(Cl_(2))+V_(Cl_(2))` (produced)`+V_(O_(2))=75mL`
`V_(Cl_(2))` (produced) `+V_(O_(2))=75-30=45mL`
(Since `V_(Cl_(2))` initiall `=30mL`)
NaOH apart from `CO_(2)` also absorbs `Cl_(2)`. So after `NaOH` treatment, the residual volume corresponds to the volume of `O_(2)(=15mL)`
`therefore` Volume of `Cl_(2)=45-15=30mL`
Volume of `Cl_(2)=30xx(x)/(2)=30impliesx=2`
Volume of `O_(2)=30xx(y)/(2)=15impliesy=1`
Hence formula of chlorid is `Cl_(2)O`.
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