The percentage of element M is 53 in its oxide of molecular formula M2O3. Its atomic mass is about

Question

The percentage of element M is 53 in its oxide of molecular formula M2O3. Its atomic mass is about (A) 45 (B) 9 (C) 18 (D) 22

Tardigrade Admin · Accepted Answer

If m is the atomic mass of the element M, then 2m + 48 g of M2O3 contain O=48 g ∴ % of O in M2O3=(48/2m+48)×100=53 (Given) ∴ 2m+48=(4800/53)=90.56 or 2m=42.56 or m=21.28=22

Q. The percentage of element M is 53 in its oxide of molecular formula $M_{2} O_{3}$ . Its atomic mass is about

45

9

18

22

If $m$ is the atomic mass of the element $M$ , then
$2 m + 48 g$ of $M_{2} O_{3}$ contain $O = 48 g$
$∴$ $%$ of $O$ in $M_{2} O_{3} = \frac{48}{2 m + 48} \times 100 = 53$
(Given)
$∴ 2 m + 48 = \frac{4800}{53} = 90.56$
or $2 m = 42.56$ or $m = 21.28 = 22$

Q. The percentage of element M is 53 in its oxide of molecular formula M2​O3​. Its atomic mass is about

45

9

18

22

If m is the atomic mass of the element M, then 2m+48g of M2​O3​ contain O=48g ∴ % of O in M2​O3​=2m+4848​×100=53 (Given) ∴2m+48=534800​=90.56 or 2m=42.56 or m=21.28=22

Q. The percentage of element M is 53 in its oxide of molecular formula $M_{2} O_{3}$ . Its atomic mass is about

If $m$ is the atomic mass of the element $M$ , then
$2 m + 48 g$ of $M_{2} O_{3}$ contain $O = 48 g$
$∴$ $%$ of $O$ in $M_{2} O_{3} = \frac{48}{2 m + 48} \times 100 = 53$
(Given)
$∴ 2 m + 48 = \frac{4800}{53} = 90.56$
or $2 m = 42.56$ or $m = 21.28 = 22$