1.

A compound of sodium does not give CO_(2) when heated but it gives CO_(2) when treated with dilute acids. A crystalline compound is found to have 37.1% Na and 14.52% H_(2)O. Hence, compound is

Answer»

`NaHCO_(3)*10H_(2)O`
`NaHCO_(3)*5H_(2)O`
`Na_(2)CO_(3)*10H_(2)O`
`Na_(2)CO_(3)*H_(2)O`

Solution :`CO_(2)` is obatined only when the compound is decomposed by acid thus, it is `Na_(2)CO_(3)`. It cannot be `NaHCO_(3)` since, it given `CO_(2)` on heating.
Probable compound is `Na_(2)CO_(3).xH_(2)O`
`{:(,,"MOLAR RATIO","Ratio"),(Na_(2)CO_(3),85.48,0.8064,1),(H_(2)O,14.52,0.8064,1):}`
Thus, compound is `Na_(2)CO_(3)*H_(2)O`
MOLECULAR WEIGHT `=124" g "mol^(-1)`
percentage of sodium `=(2xx23)/(124)xx100=37.1%`


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