0.05 g of a sample of `KClO_3` containing some KCl on decomposition liberated just sufficient oxygen for complete oxidation of 20 " mL of " CO. The volume of CO was measured at `27^@C` and 750 mm Hg. Calculate the perentage purity of `KClO_3`.

Explanation

The number of moles of CO at `27^@C` and 75 mm Hg pressure
`PV=nRT`
`n=(PV)/(RT)=(750)/(760)xx(20)/(1000)xx(1)/(0.082xx300)=8.023xx10^-4mol`
`therefore` number of moles `O_2=(8.023xx10^(-4))/(2)`
`=4.0115xx10^(-4)` mol
`2KCIO_3to2KCl+3O_2`
`2CO+O_2to2CO_2`
Weight of `KCIO_3` (pure) `=(4.0115xx10^(-4)xx2xx122.5)/(3)`
`=0.03276g`
`%` of purity`=(0.03276)/(0.05)xx100=65.52`