2 gm of a metal when dissolved in `HNO_(3)` , gets converted into nitrate salt. The nitrate was then precipitated to form 2.66 gm of metal chloride. Equivalent mass of metal can be :

Options

A. 28
B. 23.5
C. 9
D. 108

Answer

To find the equivalent mass of the metal, we can follow these steps:

### Step 1: Understand the Reaction
When the metal (M) reacts with nitric acid (HNO₃), it forms a metal nitrate (M(NO₃)₂). This metal nitrate can then be converted to a metal chloride (MCl) upon precipitation.

### Step 2: Establish the Relationship
According to the principle of chemical equivalence, the gram equivalent of the metal is equal to the gram equivalent of the chloride formed. This can be expressed mathematically as:
\[ \text{Gram equivalent of metal} = \text{Gram equivalent of chloride} \]

### Step 3: Write the Equation
Let the equivalent weight of the metal be \( E \). The mass of the metal used is 2 g, so:
\[ \text{Gram equivalent of metal} = \frac{2 \text{ g}}{E} \]
The mass of the metal chloride formed is 2.66 g. The equivalent weight of chloride (Cl) is 35.5 g, so:
\[ \text{Gram equivalent of chloride} = \frac{2.66 \text{ g}}{E + 35.5} \]

### Step 4: Set Up the Equation
From the equivalence established in Step 2, we can set up the equation:
\[ \frac{2}{E} = \frac{2.66}{E + 35.5} \]

### Step 5: Cross Multiply
Cross-multiplying gives:
\[ 2 \times (E + 35.5) = 2.66 \times E \]

### Step 6: Simplify the Equation
Expanding the left side:
\[ 2E + 71 = 2.66E \]

Rearranging the equation:
\[ 71 = 2.66E - 2E \]
\[ 71 = 0.66E \]

### Step 7: Solve for E
Now, divide both sides by 0.66 to find \( E \):
\[ E = \frac{71}{0.66} \approx 107.58 \text{ g} \]

### Step 8: Round the Answer
Rounding gives us:
\[ E \approx 108 \text{ g} \]

### Conclusion
The equivalent mass of the metal is approximately 108 g.