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If `2.73g` of oxide of vanadium contains `1.53g` fo the metal vanadium, the empirical formula of the oxide isA. `V_(2) O_(5)`B. `V_(2) O_(3)`C. `V_(3) O_(4)`D. `V O_(2)` |
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Answer» Correct Answer - A `n_(V) = (1.53g)/(51g mol^(-1)) = 0.03, n_(O) = (1.20g)/(16g mol^(-1)) = 0.075` `(n_(V))/(n_(O)) = (0.03)/(0.075) = (30)/(75) = (2xx15)/(5xx15) = 2:5` `:. EF` is `V_(2) O_(5)`. |
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