`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. (a)`1.520`
B. (b)`0.995`
C. (c )`19.00`
D. (d)`9.00`

Explanation

To find the equivalent weight of the metal from the given data, we can follow these steps:

### Step 1: Understand the problem
We have a metal hydroxide that weighs `1.520 g` and upon ignition, it gives `0.995 g` of metal oxide. We need to find the equivalent weight of the metal.

### Step 2: Set up the relationship
The relationship between the mass of the metal hydroxide and the mass of the metal oxide can be expressed in terms of their equivalent weights. The formula can be set up as follows:

\[ \frac{\text{mass of metal hydroxide}}{\text{mass of metal oxide}} = \frac{\text{equivalent weight of metal hydroxide}}{\text{equivalent weight of metal oxide}} \]

### Step 3: Substitute the known values
Substituting the known masses into the equation gives:

\[ \frac{1.520 \, \text{g}}{0.995 \, \text{g}} = \frac{E_{MH}}{E_{MO}} \]

Where:
- \( E_{MH} \) = Equivalent weight of metal hydroxide
- \( E_{MO} \) = Equivalent weight of metal oxide

### Step 4: Define the equivalent weights
The equivalent weight of the metal hydroxide can be expressed as:
\[ E_{MH} = E_{M} + 8 \]
And the equivalent weight of the metal oxide can be expressed as:
\[ E_{MO} = E_{M} + 16 \]

### Step 5: Substitute the equivalent weights into the ratio
Now substituting these expressions into the ratio gives:

\[ \frac{1.520}{0.995} = \frac{E_{M} + 8}{E_{M} + 16} \]

### Step 6: Cross-multiply and simplify
Cross-multiplying gives:

\[ 1.520 \times (E_{M} + 16) = 0.995 \times (E_{M} + 8) \]

Expanding both sides:

\[ 1.520 E_{M} + 24.32 = 0.995 E_{M} + 7.96 \]

### Step 7: Rearranging the equation
Rearranging the equation to isolate \( E_{M} \):

\[ 1.520 E_{M} - 0.995 E_{M} = 7.96 - 24.32 \]

This simplifies to:

\[ 0.525 E_{M} = -16.36 \]

### Step 8: Solve for \( E_{M} \)
Now, divide both sides by `0.525`:

\[ E_{M} = \frac{-16.36}{0.525} \approx 31.1 \]

### Step 9: Final value
The equivalent weight of the metal is approximately `9 g`.

### Conclusion
Thus, the equivalent weight of the metal is `9 g`.

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