$$5.0$$ $$cm3$$ of $$H2O2$$ liberates $$0.508$$ g of iodine from an acidified KI solution. The strength of $$H2O2$$ solution in terms of volume strength at STP is

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$$VH2O2=5ml$$ $$WI2=0.508gr$$ $$MWI2=2$$×$$127=254gr$$ $$H2O2+2KI$$→$$H+I2+2KOH$$ $$1mole$$          →$$1mole$$  ??                → $$2$$×$$10-3$$ mole $$2$$×$$10-3$$×$$11$$ $$M=2$$×$$10-3$$×$$10005$$ $$M=0.4MFor$$ $$H2O210$$ vol.strength $$=$$ $$=0.893M$$ $$=$$ $$1.786N$$ $$=$$ $$3.03$$%(W/V)?? $$-$$  …                                              $$=$$ $$0.4M$$    $$=$$ $$4.47$$ vol

5.0 cm³ of H₂O₂ liberates 0.508 g of iodine from an acidified KI solution. The strength of H₂O₂ solution in terms of volume strength at STP is

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5.0 cm3 of H2O2 liberates 0.508 g of iodine from an acidified KI solution. The strength of H2O2 solution in terms of volume strength at STP is

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5.0 cm³ of H₂O₂ liberates 0.508 g of iodine from an acidified KI solution. The strength of H₂O₂ solution in terms of volume strength at STP is