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)

• A. $\frac{X}{28.83(1-X)}$ = m (D)
• B. $\frac{1-X}{(X)^2}$ = m (C)

Answer 1

Answer: (B)

$\frac{1-X}{(X)^2}$ = m (C)

Answer 2

The problem asks for the new molarity of a H₂O₂ solution after some decomposition.

1. Initial Molarity of H₂O₂ Solution:

The "volume strength" of H₂O₂ solution indicates the volume of oxygen gas (at STP, i.e., 273 K and 1 atm) that 1 liter of the H₂O₂ solution will produce upon decomposition.

The decomposition reaction is:

$2\text{H}_2\text{O}_2(\text{aq}) \rightarrow 2\text{H}_2\text{O}(\text{l}) + \text{O}_2(\text{g})$

From the stoichiometry, 2 moles of H₂O₂ produce 1 mole of O₂.
At STP, 1 mole of O₂ occupies 22.4 liters.
Therefore, 2 moles of H₂O₂ produce 22.4 liters of O₂.
This means 1 mole of H₂O₂ produces 11.2 liters of O₂.

The relationship between volume strength (V) and Molarity (M) of H₂O₂ is:

$M = \frac{\text{Volume Strength}}{11.2}$

Given initial volume strength = 45.4 V.
Initial Molarity ($M_{\text{initial}}$) = $\frac{45.4}{11.2} \text{ M}$
$M_{\text{initial}} \approx 4.0536 \text{ M}$

2. Initial Moles of H₂O₂:

The initial volume of the H₂O₂ solution is 500 ml = 0.5 L.
Initial moles of H₂O₂ ($n_{\text{initial}}$) = $M_{\text{initial}} \times \text{Volume (L)}$
$n_{\text{initial}} = \frac{45.4}{11.2} \times 0.5 = \frac{45.4}{22.4} \text{ moles}$
$n_{\text{initial}} \approx 2.0268 \text{ moles}$

3. Moles of O₂ Lost:

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

4. Moles of H₂O₂ Decomposed:

From the decomposition reaction ($2\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2\text{O} + \text{O}_2$), 1 mole of O₂ is produced from 2 moles of H₂O₂.
Moles of H₂O₂ decomposed = $2 \times \text{Moles of O₂ lost}$
Moles of H₂O₂ decomposed = $2 \times 0.5 \text{ moles} = 1.0 \text{ mole}$

5. Final Moles of H₂O₂:

Final moles of H₂O₂ ($n_{\text{final}}$) = Initial moles of H₂O₂ - Moles of H₂O₂ decomposed
$n_{\text{final}} = \frac{45.4}{22.4} - 1.0 = \frac{45.4 - 22.4}{22.4} = \frac{23}{22.4} \text{ moles}$
$n_{\text{final}} \approx 1.0268 \text{ moles}$

6. New Molarity of H₂O₂ Solution:

The problem states to assume no change in volume, so the volume of the solution remains 0.5 L.
New Molarity ($M_{\text{new}}$) = $\frac{\text{Final moles of H₂O₂}}{\text{Volume of solution (L)}}$
$M_{\text{new}} = \frac{23/22.4}{0.5} = \frac{23}{22.4 \times 0.5} = \frac{23}{11.2} \text{ M}$
$M_{\text{new}} \approx 2.0536 \text{ M}$

The options provided are in terms of 'X' and 'm'. Let's assume 'm' is the new molarity and 'X' is the fraction of H₂O₂ decomposed.
Fraction of H₂O₂ decomposed (X) = $\frac{\text{Moles of H₂O₂ decomposed}}{\text{Initial moles of H₂O₂}}$
$X = \frac{1.0}{45.4/22.4} = \frac{22.4}{45.4}$

Let's check option (C): $m = \frac{1-X}{(X)^2}$
Substitute $X = \frac{22.4}{45.4}$:
$m = \frac{1 - \frac{22.4}{45.4}}{(\frac{22.4}{45.4})^2} = \frac{\frac{45.4 - 22.4}{45.4}}{(\frac{22.4}{45.4})^2} = \frac{\frac{23}{45.4}}{\frac{22.4^2}{45.4^2}}$
$m = \frac{23}{45.4} \times \frac{45.4^2}{22.4^2} = \frac{23 \times 45.4}{22.4^2} = \frac{1044.2}{501.76} \approx 2.081 \text{ M}$

Our calculated value is $\frac{23}{11.2} \approx 2.0536 \text{ M}$.

The value from option (C) (2.081 M) is very close to our calculated value (2.0536 M). The slight discrepancy might be due to rounding in the option's derivation or the use of slightly different constants. Given the options are formulas, and (C) yields a numerically close result, it is the most plausible choice if one must be selected. However, based on direct calculation, the new molarity is 2.0536 M.

Since the question asks for the new molarity and provides options in a formulaic form, and option (C) provides a value numerically close to the calculated one when X is defined as the fraction decomposed, we select (C).