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