Equivalent weights of two oxides of an element are 14 and 11 respectively. What is the ratio of atomicity of oxygen in the second oxide to first oxide?

Answer

To solve the problem, we need to find the ratio of atomicity of oxygen in the second oxide to that in the first oxide. Let's break down the steps systematically.

### Step 1: Define the Oxides
Let the two oxides of the element be represented as \( X_2O_p \) and \( X_2O_q \), where:
- \( p \) is the number of oxygen atoms in the first oxide.
- \( q \) is the number of oxygen atoms in the second oxide.

### Step 2: Write the Molecular Mass Expressions
The molecular mass of the first oxide \( X_2O_p \) can be expressed as:
\[ \text{Molecular mass of } X_2O_p = 2m + 16p \]
where \( m \) is the atomic mass of element \( X \) and 16 is the atomic mass of oxygen.

The molecular mass of the second oxide \( X_2O_q \) can be expressed as:
\[ \text{Molecular mass of } X_2O_q = 2m + 16q \]

### Step 3: Write the Equivalent Mass Expressions
The equivalent mass of the first oxide is given as 14:
\[ \text{Equivalent mass} = \frac{\text{Molecular mass}}{n \text{ factor}} = \frac{2m + 16p}{2p} = 14 \]
From this, we can rearrange to find:
\[ 2m + 16p = 28p \quad \Rightarrow \quad 2m = 28p - 16p \quad \Rightarrow \quad 2m = 12p \quad \Rightarrow \quad m = 6p \]

For the second oxide, the equivalent mass is given as 11:
\[ \text{Equivalent mass} = \frac{2m + 16q}{2q} = 11 \]
Rearranging gives:
\[ 2m + 16q = 22q \quad \Rightarrow \quad 2m = 22q - 16q \quad \Rightarrow \quad 2m = 6q \quad \Rightarrow \quad m = 3q \]

### Step 4: Equate the Two Expressions for \( m \)
Now we have two expressions for \( m \):
1. \( m = 6p \)
2. \( m = 3q \)

Setting them equal to each other:
\[ 6p = 3q \quad \Rightarrow \quad \frac{q}{p} = 2 \quad \Rightarrow \quad q = 2p \]

### Step 5: Find the Ratio of Atomicity
The ratio of atomicity of oxygen in the second oxide to that in the first oxide is:
\[ \frac{q}{p} = \frac{2p}{p} = 2 \]

### Final Answer
Thus, the ratio of atomicity of oxygen in the second oxide to that in the first oxide is:
\[ \text{Ratio} = 2:1 \]