MCQ
On reduction with hydrogen, $3.6 \,g$ of an oxide of metal left $3.2 \,g$ of metal. If the vapour density of metal is $32$, the simplest formula of the oxide would be
- A$MO$
- B${M_2}{O_3}$
- ✓${M_2}O$
- D${M_2}{O_5}$
✓
Answer
Correct option: C.
${M_2}O$
c
(c) As we know that
(c) As we know that
Equivalent weight $ = \frac{{{\rm{weight \,of \,metal}}}}{{{\rm{weight \,of \,oxygen}}}} \times 8$ $ = \frac{{32}}{{0.4}} \times 8 = 64$
Vapour density $ = \frac{{{\rm{mol}}{\rm{. \,wt}}}}{{\rm{2}}}$
Mol. wt $ = 2 \times V.D = 2 \times 32 = 64$
As we know that $n = \frac{{{\rm{mol\,}}{\rm{.}}\,{\rm{wt}}}}{{{\rm{\,eq}}{\rm{. wt}}}} = \frac{{64}}{{64}} = 1$
Suppose, the formula of metal oxide be ${M_2}{O_n}$. Hence the formula of metal oxide $ = {M_2}O$.
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