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On heating 1.763 g of hydrated BaCl(2) t...

On heating `1.763 g` of hydrated `BaCl_(2)` to dryness, `1.505 g` of anhyrous salt remained, What is the formula of hydrate?

Text Solution

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Let the formula of hydrate `= BaCl_(2). nH_(2) O`
`BaCl_(2). nH_(2)O = (137 + 2 xx 35.5 + 18n) = (208 + 18n) g`
`BaCl_(2) = 137 + 71 = 208 g`
`BaCl_(2). nH_(2) O rarr BaCl_(2) + nH_(2) O`
`:. (208 + 18n) g implies 208 g "of" BaCl_(2)`
`1.763 g implies (208 xx 1.763)/((208 + 18n)) = 1.505`
Solve for `n`.
`n = 1.98 ~~2`
Formula `implies BaCl_(2). 2H_(2) O`
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