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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
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The correct answer is:
100
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 $=100 \mathrm{u}$
$\mathrm{M}_2 \mathrm{O}_x$ 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 $=100 \mathrm{u}$
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