`1.520 g` of the hydroxide of a metal on ignition gave `0.995 g` of oxide. The equivalent weight of metal is

Options

A. 1.52 g
B. 0.995 g
C. 19 g
D. 9 g

Answer

To find the equivalent weight of the metal, we will follow these steps:

### Step 1: Understand the relationship between hydroxide and oxide
When the metal hydroxide is ignited, it decomposes to form the metal oxide and water. We can set up a relationship between the weights of the metal hydroxide and the metal oxide based on their equivalent weights.

### Step 2: Write the equation for the relationship
The relationship can be expressed as:
\[ \frac{\text{Weight of Metal Hydroxide}}{\text{Weight of Metal Oxide}} = \frac{\text{Equivalent Weight of Metal Hydroxide}}{\text{Equivalent Weight of Metal Oxide}} \]

### Step 3: Assign known values
From the problem, we have:
- Weight of metal hydroxide = \(1.520 \, \text{g}\)
- Weight of metal oxide = \(0.995 \, \text{g}\)

### Step 4: Define the equivalent weights
Let:
- \(x\) = equivalent weight of the metal
- Equivalent weight of hydroxide ion (OH\(^-\)) = \(\frac{17}{1} = 17\)
- Equivalent weight of metal hydroxide (M(OH)\(_n\)) = \(x + 17\)

For the metal oxide (MO):
- Equivalent weight of oxide ion (O\(^{2-}\)) = \(\frac{16}{2} = 8\)
- Equivalent weight of metal oxide (MO) = \(x + 8\)

### Step 5: Set up the equation
Now substituting into the equation:
\[ \frac{1.520}{0.995} = \frac{x + 17}{x + 8} \]

### Step 6: Cross-multiply and simplify
Cross-multiplying gives:
\[ 1.520(x + 8) = 0.995(x + 17) \]

Expanding both sides:
\[ 1.520x + 12.16 = 0.995x + 16.865 \]

### Step 7: Rearranging the equation
Rearranging to isolate \(x\):
\[ 1.520x - 0.995x = 16.865 - 12.16 \]
\[ 0.525x = 4.705 \]

### Step 8: Solve for \(x\)
Now, divide both sides by \(0.525\):
\[ x = \frac{4.705}{0.525} \approx 8.95 \approx 9.0 \]

### Conclusion
The equivalent weight of the metal is approximately \(9.0 \, \text{g/equiv}\).

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