In an experiment, 4 g of M2Ox oxide was reduced to 2.8 g of the metal. If the atomic mass of the metal is 56 g mol-1, the number of O atoms in the oxide is

Question

In an experiment, 4 g of M2Ox oxide was reduced to 2.8 g of the metal. If the atomic mass of the metal is 56 g mol-1, the number of O atoms in the oxide is (A) 1 (B) 2 (C) 3 (D) 4

Tardigrade Admin · Accepted Answer

Mass of oxygen in oxide =4-2.8 =1.2 g Eq. wt. of metal =( text mass of metal / text mass of oxygen ) × 8 =(2.8/1.2) × 8 =18.67 Valency of metal =( text atomic wt. of metal / text eq. wt. of metal ) =(56/18.67) (given) =3

Q. In an experiment, $4 g$ of $M_{2} O_{x}$ oxide was reduced to $2.8 g$ of the metal. If the atomic mass of the metal is $56 g m o l^{- 1},$ the number of $O$ atoms in the oxide is

1

2

3

4

Mass of oxygen in oxide $= 4 - 2.8$
$= 1.2 g$
Eq. wt. of metal $= \frac{mass of metal}{mass of oxygen} \times 8$ &
$= \frac{2.8}{1.2} \times 8$
$= 18.67$
Valency of metal $= \frac{atomic wt. of metal}{eq. wt. of metal}$
$= \frac{56}{18.67}$ (given)
$= 3$

Q. In an experiment, 4g of M2​Ox​ oxide was reduced to 2.8g of the metal. If the atomic mass of the metal is 56gmol−1, the number of O atoms in the oxide is

1

2

3

4

Mass of oxygen in oxide =4−2.8 =1.2g Eq. wt. of metal = mass of oxygen mass of metal ​×8& =1.22.8​×8 =18.67 Valency of metal = eq. wt. of metal atomic wt. of metal ​ =18.6756​ (given) =3

Q. In an experiment, $4 g$ of $M_{2} O_{x}$ oxide was reduced to $2.8 g$ of the metal. If the atomic mass of the metal is $56 g m o l^{- 1},$ the number of $O$ atoms in the oxide is

Mass of oxygen in oxide $= 4 - 2.8$
$= 1.2 g$
Eq. wt. of metal $= \frac{mass of metal}{mass of oxygen} \times 8$ &
$= \frac{2.8}{1.2} \times 8$
$= 18.67$
Valency of metal $= \frac{atomic wt. of metal}{eq. wt. of metal}$
$= \frac{56}{18.67}$ (given)
$= 3$