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If 2.73g of oxide of vanadium contains 1...

If `2.73g` of oxide of vanadium contains `1.53g` fo the metal vanadium, the empirical formula of the oxide is

A

`V_(2) O_(5)`

B

`V_(2) O_(3)`

C

`V_(3) O_(4)`

D

`V O_(2)`

Text Solution

Verified by Experts

The correct Answer is:
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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