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

Question

The percentage of an element M is 53 in its oxide of molecular formula M2 O3. Its atomic mass is about (A) 45 (B) 12 (C) 18 (D) 27

Tardigrade Admin · Accepted Answer

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 M2 O3=(48/2 m+48) × 100=53 2 m+48=(4800/53)=90.56 2 m=42.56 m=26.58=27

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

45

12

18

27

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 $M_{2} O_{3} = \frac{48}{2 m + 48} \times 100 = 53$
$2 m + 48 = \frac{4800}{53} = 90.56$
$2 m = 42.56$
$m = 26.58 = 27$

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

45

12

18

27

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 2m+48=534800​=90.56 2m=42.56 m=26.58=27

Q. The percentage of an 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 $M_{2} O_{3} = \frac{48}{2 m + 48} \times 100 = 53$
$2 m + 48 = \frac{4800}{53} = 90.56$
$2 m = 42.56$
$m = 26.58 = 27$