In an experiment, 4g of `M_(2)O_(x)` oxide was reduced to 2.8g of the metal. If the atomic mass of the metal is `56g"mol"^(-1)`, the number of oxygen atoms in the oxide is:A. 1B. 2C. 3D. 4

In an experiment, 4g of `M_(2)O_(x)` oxide was reduced to 2.8g of the

1.	In an experiment, 4g of `M_(2)O_(x)` oxide was reduced to 2.8g of the metal. If the atomic mass of the metal is `56g"mol"^(-1)`, the number of oxygen atoms in the oxide is:A. 1B. 2C. 3D. 4
Answer» Correct Answer - C `M_(2)O_(x) overset("Reduction")(rarr) M` Eq. of `M_(2)O_(x)=` eq. of Metal `("wt. of "M_(2)O_(x))/("Eq. wt. of "M_(2)O_(x))=("Wt. of Metal")/("Eq. wt. of Metal")` `4/((2xx56+xxx16)/(2x))=2.8/(56/x)` ...(i) On solving we get, `4/(56+8x)=2.8/56 implies 1/(14+2x)=1/20 implies 2x=6 implies x=3` Hence, the oxide is `M_(2)O_(3)`.

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In an experiment, 4g of `M_(2)O_(x)` oxide was reduced to 2.8g of the metal. If the atomic mass of the metal is `56g"mol"^(-1)`, the number of oxygen atoms in the oxide is:A. 1B. 2C. 3D. 4

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