Question

45.4 V H₂O₂ solution (500 ml) when exposed to atmosphere looses 11.2 litre of O₂ at 1 atm, & 273 K. New molarity of H₂O₂ solution (Assume no change in volume) CT0031

Answer 1

2.05 M

Answer 2

The problem asks for the new molarity of an H₂O₂ solution after it loses a certain volume of O₂ due to decomposition.

1. Initial Molarity of H₂O₂ Solution: The volume strength of H₂O₂ solution is given as 45.4 V. This means that 1 litre of this H₂O₂ solution will produce 45.4 litres of O₂ gas at STP (1 atm, 273 K). The decomposition of hydrogen peroxide is represented by the equation: $2H_2O_2(aq) \to 2H_2O(l) + O_2(g)$ From the stoichiometry, 2 moles of H₂O₂ produce 1 mole of O₂. At STP, 1 mole of O₂ occupies 22.4 litres. Therefore, 2 moles of H₂O₂ produce 22.4 litres of O₂ at STP. This implies that 1 mole of H₂O₂ produces 11.2 litres of O₂ at STP.

The molarity (M) of H₂O₂ solution can be calculated from its volume strength (V) using the formula: $M = \frac{\text{Volume Strength}}{11.2}$ Initial Molarity ($M_{initial}$) = $\frac{45.4}{11.2} = 4.05357 \text{ M}$

2. Initial Moles of H₂O₂ in the Solution: The volume of the H₂O₂ solution is 500 ml = 0.5 L. Initial moles of H₂O₂ ($n_{initial}$) = Molarity × Volume of solution (in L) $n_{initial} = 4.05357 \text{ M} \times 0.5 \text{ L} = 2.026785 \text{ moles}$

3. Moles of O₂ Lost: The solution loses 11.2 litres of O₂ at 1 atm and 273 K (STP conditions). Moles of O₂ lost ($n_{O_2 \text{ lost}}$) = $\frac{\text{Volume of O₂ lost}}{\text{Molar volume at STP}}$ $n_{O_2 \text{ lost}} = \frac{11.2 \text{ L}}{22.4 \text{ L/mol}} = 0.5 \text{ moles}$

4. Moles of H₂O₂ Decomposed: From the decomposition reaction ($2H_2O_2 \to 2H_2O + O_2$), 1 mole of O₂ is produced from 2 moles of H₂O₂. So, the moles of H₂O₂ decomposed ($n_{decomposed}$) = $2 \times n_{O_2 \text{ lost}}$ $n_{decomposed} = 2 \times 0.5 \text{ moles} = 1.0 \text{ mole}$

5. Remaining Moles of H₂O₂: Remaining moles of H₂O₂ ($n_{final}$) = Initial moles - Moles decomposed $n_{final} = 2.026785 \text{ moles} - 1.0 \text{ mole} = 1.026785 \text{ moles}$

6. New Molarity of H₂O₂ Solution: The problem states to assume no change in volume, so the volume of the solution remains 500 mL (0.5 L). New Molarity ($M_{new}$) = $\frac{\text{Remaining moles of H₂O₂}}{\text{Volume of solution (in L)}}$ $M_{new} = \frac{1.026785 \text{ moles}}{0.5 \text{ L}} = 2.05357 \text{ M}$

Rounding to three significant figures (consistent with the given 45.4 V): $M_{new} \approx 2.05 \text{ M}$

The final answer is $\boxed{\text{2.05 M}}$

Explanation of the solution:

1. Calculated initial molarity of H₂O₂ using the given volume strength (45.4 V) and the conversion factor (Molarity = Volume Strength / 11.2).
2. Determined the initial moles of H₂O₂ in the 500 ml solution.
3. Calculated the moles of O₂ lost from the solution using the given volume (11.2 L) and molar volume at STP (22.4 L/mol).
4. Used the stoichiometry of H₂O₂ decomposition ($2H_2O_2 \to O_2$) to find the moles of H₂O₂ that decomposed to produce the lost O₂.
5. Subtracted the decomposed moles from the initial moles to find the remaining moles of H₂O₂.
6. Calculated the new molarity by dividing the remaining moles by the original volume of the solution (assuming no volume change).