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On reduction with hydrogen, `3.6 g` of an oxide of matel left `3.2 g` of metal. If the vapour density of metal is `32`, the simplest formula of the oxide would beA. (a)`MO`B. (b)`M_(2)O_(3)`C. (c )`M_(2)O`D. (d)`M_(2)O_(5)` |
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Answer» Correct Answer - C As we know that Equivalent weight`=("weight of metal")/("weight of oxygen")xx8` `=32/0.4xx8=64` Vapour density`=(mol. Wt)/2` Mol. Wt.`=2xxV.D=2xx32=54` As we know that `n=("mol. wt")/("eq. wt")=64/64=1` Suppose, the formula of metal oxide be. `M_(2)O_(n)` Hence the formula of metal oxide`=M_(2)O`. |
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