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 solve the problem step by step, we will follow the procedure outlined in the video transcript and derive the equivalent mass of the metal.

### Step 1: Understand the Reaction
When 2 g of a metal (M) is dissolved in nitric acid (HNO₃), it forms a metal nitrate (M(NO₃)₂). This nitrate is then converted into a metal chloride (MCl₂) during precipitation.

### Step 2: Write the Equivalent Mass Relationship
According to the principle of chemical equivalency, the gram equivalent of the metal will be equal to the gram equivalent of the metal chloride formed.

Let:
- Equivalent weight of the metal = E
- Equivalent weight of metal chloride = E + Equivalent weight of chloride

### Step 3: Calculate the Equivalent Weight of Chloride
The equivalent weight of chlorine (Cl) can be calculated as:
\[ \text{Equivalent weight of Cl} = \frac{\text{Atomic mass of Cl}}{\text{Valency of Cl}} \]
The atomic mass of chlorine is approximately 35.5 g, and its valency is 1.
Thus,
\[ \text{Equivalent weight of Cl} = \frac{35.5}{1} = 35.5 \, \text{g} \]

### Step 4: Set Up the Equation
Using the equivalency principle:
\[ \frac{2 \, \text{g}}{E} = \frac{2.66 \, \text{g}}{E + 35.5} \]

### Step 5: Cross-Multiply to Solve for E
Cross-multiplying gives:
\[ 2.66 \cdot E = 2 \cdot (E + 35.5) \]
Expanding this:
\[ 2.66E = 2E + 71 \]

### Step 6: Rearranging the Equation
Rearranging the equation gives:
\[ 2.66E - 2E = 71 \]
\[ 0.66E = 71 \]

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

### Step 8: Conclusion
The equivalent mass of the metal is approximately 107.58 g. The closest option available is 108 g.

### Final Answer
The equivalent mass of the metal can be: **108 g**