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For BaCl_(2) xH_(2)O, if 2.1 gm of compound gives 2 gm of anhydrous BaSO_(4) upon treatement with H_(2)SO_(4). Then calculate value of 'x'

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Solution :POAc on Ba
`n_(BaCl_(2)).xxH_(2)O=n_(BaAO_(4))`
`(2.1)/(208+18x)=(2)/(233)`
`48.3=416+36xrArr75.3=36x`
x=2.03 LTBR. `xapprox2`


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