The maximum number of molecules is present in;

To find which sample contains the greatest number of molecules we convert each quantity to moles, then multiply by Avogadro’s number ($6.022 \times 1023 mol‑1$).

**Option A – 5 g O₂**
Molar mass of O₂ = 32 g mol^‑1.
$n\; =\; \backslash frac\{5\backslash ;\backslash text\{g\}\}\{32\backslash ;\backslash text\{g mol\}^\{-1\}\}\; =\; 0.15625\backslash ;\backslash text\{mol\}$
Number of molecules = $0.15625\; \backslash times\; 6.022 \times 10^\{23\}\; \backslash approx\; 9.4 \times 10^\{22\}$ molecules.

**Option B – 1.5 g H₂** (the answer we need to justify)
Molar mass of H₂ = 2 g mol^‑1.
$n\; =\; \backslash frac\{1.5\backslash ;\backslash text\{g\}\}\{2\backslash ;\backslash text\{g mol\}^\{-1\}\}\; =\; 0.75\backslash ;\backslash text\{mol\}$
Number of molecules = $0.75\; \backslash times\; 6.022 \times 10^\{23\}\; \backslash approx\; 4.5 \times 10^\{23\}$ molecules.

**Option C – 5 L N₂ gas at STP**
At STP, 1 mol of any ideal gas occupies 22.4 L.
$n\; =\; \backslash frac\{5\backslash ;\backslash text\{L\}\}\{22.4\backslash ;\backslash text\{L mol\}^\{-1\}\}\; =\; 0.2232\backslash ;\backslash text\{mol\}$
Number of molecules = $0.2232\; \backslash times\; 6.022 \times 10^\{23\}\; \backslash approx\; 1.35 \times 10^\{23\}$ molecules.

**Option D – 15 L H₂ gas at STP**
$n\; =\; \backslash frac\{15\backslash ;\backslash text\{L\}\}\{22.4\backslash ;\backslash text\{L mol\}^\{-1\}\}\; =\; 0.6696\backslash ;\backslash text\{mol\}$
Number of molecules = $0.6696\; \backslash times\; 6.022 \times 10^\{23\}\; \backslash approx\; 4.03 \times 10^\{23\}$ molecules.

Comparing the calculated molecule counts:
- Option A: ~9.4 × 10²²
- Option B: ~4.5 × 10²³
- Option C: ~1.35 × 10²³
- Option D: ~4.03 × 10²³

The largest value is obtained for **Option B**, 1.5 g of H₂, which corresponds to 0.75 mol or about 4.5 × 10²³ molecules. Therefore, the maximum number of molecules is present in 1.5 g of H₂ gas.