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(i)What mass of hydrogen peroxide will be present in 2 L of a 5 molar solution ? (ii) Calculate the mass of oxygen which will be liberated by the decomposition of 200 ml of this solution. |
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Answer» Solution :(i)Mol mass of `H_2O_2=34 "g mol"^(-1)` `therefore` 1L of 5 M solution of `H_2O_2` will CONTAIN 34 x 5 g `H_2O_2` 2L of 5 M solution of `H_2O_2` will contain 34 x 2 x 5=340 g `H_2O_2` (II) 0.2 L or 200 ML of 5 M solution will contain `H_2O_2` `=(340xx0.2)/2`=34 g `H_2O_2` Now, `{:(2H_2O_2 to , 2H_2O + O_2),(2 XX 34 = 68 gm , 2 xx 16 =32 gm):}` Now, 68 g of `H_2O_2` on decomposition will give `O_2` =32 g 34 g of `H_2O_2` on decomposition will give `=(32xx34)/68`=16 g `O_2`. |
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