Essential quantity of ammonium sulphate taken for preparation of 1 molar sollution in 2 litres is

Options

A. 132 gm
B. 264 gm
C. 198 gm
D. 212 gm

Explanation

To determine the essential quantity of ammonium sulfate needed to prepare a 1 molar solution in 2 liters, follow these steps:

### Step 1: Understand Molarity
Molarity (M) is defined as the number of moles of solute per liter of solution. Therefore, for a 1 molar solution, we need 1 mole of solute for every liter of solution.

### Step 2: Calculate the Number of Moles Required
Since we want to prepare 2 liters of a 1 molar solution:
\[ \text{Number of moles} = \text{Molarity} \times \text{Volume (in liters)} = 1 \, \text{mol/L} \times 2 \, \text{L} = 2 \, \text{moles} \]

### Step 3: Find the Molecular Weight of Ammonium Sulfate
The formula for ammonium sulfate is \( (NH_4)_2SO_4 \). We need to calculate its molecular weight:
- Nitrogen (N): 14 g/mol (2 Nitrogens = 2 × 14 = 28 g)
- Hydrogen (H): 1 g/mol (8 Hydrogens = 8 × 1 = 8 g)
- Sulfur (S): 32 g/mol (1 Sulfur = 32 g)
- Oxygen (O): 16 g/mol (4 Oxygens = 4 × 16 = 64 g)

Now, add these together:
\[ \text{Molecular weight of } (NH_4)_2SO_4 = 28 + 8 + 32 + 64 = 132 \, \text{g/mol} \]

### Step 4: Calculate the Weight of Ammonium Sulfate Needed
Now that we know we need 2 moles of ammonium sulfate, we can calculate the weight:
\[ \text{Weight} = \text{Number of moles} \times \text{Molecular weight} = 2 \, \text{moles} \times 132 \, \text{g/mol} = 264 \, \text{grams} \]

### Final Answer
The essential quantity of ammonium sulfate needed to prepare a 1 molar solution in 2 liters is **264 grams**.

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