A metal (M) forms two oxides. The ratio M:O ( | vedclass

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(B) Let the two oxides be $M_x O_y$ and $M_x O_z$. According to the problem, the ratio of mass of $M$ to mass of $O$ is given as $\frac{M_{mass}}{O_{m

Vedclass · Accepted Answer

$100$

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(B) Let the two oxides be $M_x O_y$ and $M_x O_z$. According to the problem, the ratio of mass of $M$ to mass of $O$ is given as $\frac{M_{mass}}{O_{mass}} = \frac{25}{4}$ and $\frac{25}{6}$. Let $A$ be the atomic mass of metal $M$. For the first oxide: $\frac{x 	imes A}{y 	imes 16} = \frac{25}{4} \implies \frac{x 	imes A}{y} = 100$. For the second oxide: $\frac{x 	imes A}{z 	imes 16} = \frac{25}{6} \implies \frac{x 	imes A}{z} = \frac{200}{3}$. From the first equation, $x 	imes A = 100y$. Substituting this into the second equation: $\frac{100y}{z} = \frac{200}{3} \implies \frac{y}{z} = \frac{2}{3}$. Thus, $y=2$ and $z=3$ are the smallest integers. Substituting $y=2$ into $x 	imes A = 100 	imes 2 = 200$. For minimum atomic mass, we take $x=2$, so $2 	imes A = 200 \implies A = 100 \ u$.

$A$ 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 mass of $M$ is

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