Question

A metal (M) forms two oxides. The ratio M:O (by weight) in the two oxides are $25:4$ and 25:6. The minimum value of atomic of mathrm $M$ is

• A. 50
• B. 100
• C. 150
• D. 200

Answer 1

Answer: (B)

Let two oxide be $M_2{O}_x$ and $M_{2}Oy$
As per question
$\frac{2a}{16x}=\frac{25}{4}$ and $\frac{2a}{16y}=\frac{25}{6}$
$x=\frac{a}{50},y=\frac{3a}{100}$ where $a=$ atomic mass of Metal
As $x$ and $y$ to be an integer,
If we take $a=50$, then $x=1,y=1.5$ (not possible)
If we take $a=100$ then $x=2,y=3$ (possible)
$\therefore$ Minimum Atomic Mass $=100u$