To solve this problem, we need to determine the number of moles of O3 and O2 involved when 1 mole of PbS is oxidized. The chemical reaction for oxidizing PbS using O3 is:
PbS + 4O3 → PbSO4 + 4O2
Step-by-step explanation:
From the balanced reaction, 1 mole of PbS requires 4 moles of O3 to be oxidized to PbSO4.
During this process, it produces 4 moles of O2.
Given these stoichiometric relationships, the number of moles of O3 (“X” moles) used is 4, and the number of moles of O2 (“Y” moles) produced is also 4.
The problem asks for the sum X + Y, which is:
X + Y = 4 + 4 = 8
Verification:
The computed value of X + Y = 8 falls within the expected range of [8,8].
The balanced chemical equation for the oxidation of PbS by ozone is:
\(\text{PbS} + 4\text{O}_3 \rightarrow \text{PbSO}_4 + 4\text{O}_2\)
From the equation:
- 1 mole of PbS reacts with 4 moles of \( \text{O}_3 \), so \( X = 4 \).
- 4 moles of \( \text{O}_2 \) are produced, so \( Y = 4 \).
Therefore:
\(X + Y = 4 + 4 = 8\)
The Correct answer is: 8
200 cc of $x \times 10^{-3}$ M potassium dichromate is required to oxidise 750 cc of 0.6 M Mohr's salt solution in acidic medium. Here x = ______ .

A substance 'X' (1.5 g) dissolved in 150 g of a solvent 'Y' (molar mass = 300 g mol$^{-1}$) led to an elevation of the boiling point by 0.5 K. The relative lowering in the vapour pressure of the solvent 'Y' is $____________ \(\times 10^{-2}\). (nearest integer)
[Given : $K_{b}$ of the solvent = 5.0 K kg mol$^{-1}$]
Assume the solution to be dilute and no association or dissociation of X takes place in solution.
Inductance of a coil with \(10^4\) turns is \(10\,\text{mH}\) and it is connected to a DC source of \(10\,\text{V}\) with internal resistance \(10\,\Omega\). The energy density in the inductor when the current reaches \( \left(\frac{1}{e}\right) \) of its maximum value is \[ \alpha \pi \times \frac{1}{e^2}\ \text{J m}^{-3}. \] The value of \( \alpha \) is _________.
\[ (\mu_0 = 4\pi \times 10^{-7}\ \text{TmA}^{-1}) \]