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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. `MO`B. `M_(2)O_(3)`C. `M_(2)O`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` `=3.2/0.4xx8=64` Vapour density `=("mol. wt")/2` Mol. Wt`=2xxV.D=2xx32=64` 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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