What is the mass of hydrogen peroxide present in 1 litre of a 2M solution ? Calculate the volume of oxygen at S.T.P. liberated upon the complete decomposition at 100 `cm^3` of the above solution.

Explanation

To solve the question, we will follow these steps:

### Step 1: Calculate the mass of hydrogen peroxide (H2O2) in 1 litre of a 2M solution.

1. **Understanding Molarity**: A 2M solution means there are 2 moles of solute (H2O2) in 1 litre (1000 cm³) of solution.
2. **Molar Mass of H2O2**:
- The molar mass of hydrogen (H) = 1 g/mol
- The molar mass of oxygen (O) = 16 g/mol
- Therefore, the molar mass of H2O2 = (2 × 1) + (2 × 16) = 2 + 32 = 34 g/mol.
3. **Mass Calculation**:
- For 2 moles of H2O2:
- Mass = Number of moles × Molar mass = 2 moles × 34 g/mol = 68 grams.

### Step 2: Calculate the volume of oxygen liberated from 100 cm³ of the 2M solution.

1. **Volume of the solution**: We need to find the mass of H2O2 in 100 cm³ (or 100 mL) of the solution.
2. **Mass in 100 mL**:
- Since 1 litre = 1000 mL, then:
- Mass of H2O2 in 100 mL = (68 g / 1000 mL) × 100 mL = 6.8 grams.

3. **Decomposition Reaction**: The decomposition of hydrogen peroxide is given by the equation:
\[ 2H2O2 \rightarrow 2H2O + O2 \]
This means 2 moles of H2O2 produce 1 mole of O2.

4. **Moles of H2O2 in 6.8 grams**:
- Moles of H2O2 = Mass / Molar mass = 6.8 g / 34 g/mol = 0.2 moles.

5. **Moles of O2 produced**:
- From the reaction, 2 moles of H2O2 yield 1 mole of O2, so:
- Moles of O2 = 0.2 moles H2O2 × (1 mole O2 / 2 moles H2O2) = 0.1 moles O2.

6. **Volume of O2 at STP**:
- At standard temperature and pressure (STP), 1 mole of gas occupies 22.4 litres.
- Therefore, volume of O2 = 0.1 moles × 22.4 L/mole = 2.24 litres.

### Final Answers:
- The mass of hydrogen peroxide in 1 litre of a 2M solution is **68 grams**.
- The volume of oxygen liberated from 100 cm³ of this solution upon complete decomposition is **2.24 litres**.