The equivalent mass of a metal is twice to that of oxygen. How many times the weight of it's oxide is greater than the weight of metal?

Options

A. `1.5`
B. 2
C. 3
D. 4

Answer

To solve the problem step-by-step, we can follow these instructions:

### Step 1: Define the variables
Let:
- \( E_M \) = equivalent mass of the metal
- \( E_O \) = equivalent mass of oxygen
- \( W_M \) = weight of the metal
- \( W_O \) = weight of oxygen
- \( W_{MO} \) = weight of the metal oxide

### Step 2: Set up the relationship between equivalent masses
According to the problem, the equivalent mass of the metal is twice that of oxygen:
\[ E_M = 2E_O \]

### Step 3: Use the law of equivalents
According to the law of equivalents, the number of equivalents of metal is equal to the number of equivalents of oxygen:
\[ \frac{W_M}{E_M} = \frac{W_O}{E_O} \]

### Step 4: Rearrange the equation
We can rearrange this equation to express the weight of oxygen in terms of the weight of metal:
\[ \frac{W_O}{W_M} = \frac{E_M}{E_O} \]

### Step 5: Substitute the equivalent mass relationship
Substituting \( E_M = 2E_O \) into the equation:
\[ \frac{W_O}{W_M} = \frac{2E_O}{E_O} \]
This simplifies to:
\[ \frac{W_O}{W_M} = 2 \]

### Step 6: Express the weight of the oxide
The weight of the metal oxide \( W_{MO} \) is the sum of the weights of the metal and oxygen:
\[ W_{MO} = W_M + W_O \]

### Step 7: Substitute \( W_O \) into the oxide weight equation
From the previous step, we know \( W_O = 2W_M \). Substituting this into the equation for \( W_{MO} \):
\[ W_{MO} = W_M + 2W_M = 3W_M \]

### Step 8: Find the ratio of the weight of the oxide to the weight of the metal
Now, we can express how many times the weight of the oxide is greater than the weight of the metal:
\[ \frac{W_{MO}}{W_M} = \frac{3W_M}{W_M} = 3 \]

### Conclusion
Thus, the weight of the metal oxide is 3 times greater than the weight of the metal.

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