Two oxides of a X contain 50% and 40% of non-metal respectively. If the formula of the first oxide is $X O_{2}$ . Then the formula of second oxide is

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$X_{2} O_{5}$

$X O_{3}$

$X_{2} O$

$X_{2} O_{3}$

answer is C.

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Detailed Solution

1st Method:
Given, formula of first oxide is $X O_{2}$ means one mole of
$X = 2$ mole of oxygen in the first oxide.
For second oxide,
40% of $(X),$ we have 60% of oxygen i.e.,
For 40% of $(X),$ . We have $2 \times 60 %$ of oxygen.
$2 \times 60 %$ For 100% of $(X)$ (i.e., one mole) we have, $\frac{2 \times 60}{40} = 3$ moles of oxygen.
$∴$ Formula of second oxide is $= X O_{3}$
2nd Method:
In $X O_{2}, X : O : : 50 : 50$
Mass of oxygen $= 2 \times 16 = 32 g$
Hence, mass of $X = 32 g$
In $X O_{n}, X : O : : 40 : 60$
Mass of $X,$ in this case also remains $= 32 g$
$40 % \equiv 32 g$
$60 % = \frac{32}{40} \times 60 = 48 g$
Mass of oxygen $= 48 g$
$\Rightarrow n = \frac{48}{16} = 3$
Formula of oxide $= X O_{3}$