Q3 / 155 · WBJEE 2022 · Chemistry

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

  • A
    50
  • Correct answer
    B
    100
  • C
    150
  • D
    200

Solution

Let two oxide be $\mathrm{M}_2 \mathrm{O}_x$ and $\mathrm{M}_2 \mathrm{Oy}$ $\underline{\text{As per question}}$ $\frac{2 a}{16 x}=\frac{25}{4}$ and $\frac{2 a}{16 y}=\frac{25}{6}$ $x=\frac{a}{50}, y=\frac{3 a}{100}$ 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 $=100 \mathrm{u}$

Hence, the correct answer is option: 'B'.

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