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5.0 cm^(3) of H_(2)O_(2) liberates 0.508 g of iodine from an acidified KI solution. The strength of H_(2)O_(2) solution in terms of volume strenth at STP is |
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Answer» 6.48 volumes `2KI+H_(2)SO_(4)+underse(5.0 cm^(3))underset(34 g)(H_(2)O_(2))rarr K_(2)SO_(4)+2H_(2)O+underset(0.508 g)underset(254 g)(I_(2))` Then, Mass of `H_(2)O_(2)` that liberates 0.508 g of `I_(2)` `= (34)/(254)xx0.508 = 0.068 g` This much `H_(2)O_(2)` is present in `5.0 cm^(3)`. Hydrogen PEROXIDE decomposes as follows : `underset(2xx34 g=68 g)underset(2 mol)(2H_(2)O_(2))rarr 2H_(2)O underset(22400 cm^(3)"(at STP)")underset(1 mol)(+O_(2))` Then 68 g of `H_(2)O_(2)` GIVES 22400 `cm^(3)` of oxygen 1 g of `H_(2)O_(2)` gives `(22400 cm^(3))/(68)` of oxygen 0.068 g of `H_(2)O_(2)` gives `(22400 cm^(3))/(68 g)xx 0.068 g = 22.4 cm^(3)` of oxygen Thus, `5 cm^(3)` of `H_(2)O_(2)` gives `22.4 cm^(3)` of oxygen at STP So, `1 cm^(3)` of `H_(2)O_(2)` gives `(22.4)/(5)cm^(3)` of oxygen at STP `= 4.48 cm^(3)` at STP THEREFORE, strength of the given `H_(2)O_(2)` sample = 4.48 volumes. |
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