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Two oxides of a certain metal were separately heated in a current of hydrogen until constant weights were obtained. The water produced in each case was carefully collected and weighed. 2 grams of each oxide gave respectively 0.2517 grams and 0.4526 grams of water. Show that these results established the Law of Multiple Proportions. |
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Answer» Solution :Step 1. To calculate the mass of oxygen in each oxide. Here, we are given : Mass of each oxide = 2.0 g Mass of WATER produced in case I = 0.2517 g Mass of water produced in case II = 0.4526 g 18 g of water `(H_(2)O)` contain oxygen = 16 g `THEREFORE"0.2517 g of water contains oxygen "=(16)/(18)xx0.2517g=0.2237g` and 0.4526 g of water contains oxygen `=(16)/(18)xx0.4526g=0.4023g` Step 2. To calculate the mass of oxygenwhich would combine with 1 g of metal in EAH oxide. In case I. Mass of metal oxide = 2g Mass of oxygen = 0.2237 g `therefore"Mass of metal "=2-0.2237=1.7763g` `therefore" Mass of oxygen which combines with 1.7763 g of metal "=0.2237 g` `therefore"Mass of oxygen which combines with 1 g of metal"=(0.2237)/(1.7763)g=0.1259g` In case II. Mass of metal oxide = 2 g Mass of oxygen = 0.4023 g `therefore"Mass of metal"=2-0.4023=1.5977g` Mass of oxygen which combines with 1.5977 g of metal = 0.4023 g `therefore"Mass of oxygen which combines with 1 g of metal "=(0.4023)/(1.5977)g=0.2515g` Step 3. To compare the MASSES of oxygenwhich combine iwith the same mass of metal in the two oxides. The masses of oxygen which combine with 1 g of metal in the two oxides are respectively 0.1259 g and 0.2513 g. These masses are in the ratio `0.1259:0.2515 or 1:2` Since, this is a simple ratio, so the above results establish the Law of Multiple Proportions. |
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