1.

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

Answer» `underset(underset((208+18n))((137+71+18n)))(BaCl_(2).nH_(2)O)overset("Heat")(rarr)underset(underset((208))((137+71)))(BaCl_(2))`
According to available data,
`(208 + 18n)g` of `BaCl_(2).nH_(2)O` give anhydrous salt = 208 g
1.763 g of `BaCl_(2).nH_(2)O` give anhydrous salt `= ((208g))/((208+18n))xx(1.764g)`
But anhydrous `BaCl_(2)` actually formed = 1.505 g
or `((208))/((208+18n))xx(1.763g)=1.505g`
or `(208+18n)=(208xx(1.763g))/((1.505g))=243.66`
or `18n=243.66-208=35.66orn=(35.66)/(18)~~2`
`:.` Formula of hydrated salt `= BaCl_(2).2H_(2)O`.


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