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 mass of $M$ is

• A. 50
• B. 100
• C. 150
• D. 200

Answer 1

The Correct Option is B

We are given that a metal ($M$) forms two oxides, and the ratios of the metal to oxygen (by weight) in the two oxides are:

1. $25:4$ (for the first oxide) 2. $25:6$ (for the second oxide) 

Let the atomic mass of the metal be m.  

For the first oxide: The weight ratio of $M$ to $O$ is $25:4$, so: 
$\dfrac{25}{m} = \dfrac{4}{16}$ This simplifies to: 
$\dfrac{25}{m} = \dfrac{1}{4}$, so $m = 100$. 

For the second oxide: The weight ratio of $M$ to $O$ is $25:6$, so: 
$\dfrac{25}{m} = \dfrac{6}{16}$ This simplifies to: 
$\dfrac{25}{m} = \dfrac{3}{8}$, so $m = 100$. 

Thus, the minimum value of the atomic mass of $M$ is: 

Answer: 100