Maximum number of molecules are present in

No. of molecules in different cases

(1) $\because$ 22.4 litre of H₂, at STP contains = 6.023 × 10²³ molecules of H₂

${\text{∴15 litre of H}}_2 \mathrm{at} \mathrm{STP} \mathrm{contains} =\frac{15}{22.4}\times 6.023\times {10}^{23}$

= 4.03 x 10²³ molecules of H₂

(2) $\because$22.4 litre of N₂, gas at STP contains =6.023 × 10²³ molecules of N₂

$\therefore$5 litre of N₂, gas at STP contains

$=\frac{5}{22.4}\times 6.023\times {10}^{23}$

$=1.344\times {10}^{23}\text{ molecules of }{\mathrm{N}}_2$

(3) $\because 2\mathrm{g}\text{ of }{\mathrm{H}}_2=6.023\times {10}^{23}\text{ molecules of }{\mathrm{H}}_2$

$\therefore 0.5\mathrm{g}\text{ of }{\mathrm{H}}_2=\frac{0.5}{2}\times 6.023\times {10}^{23}$

$=1.505\times {10}^{23}\text{ molecules of }{\mathrm{H}}_2$

(4) Similarly 10 g of O₂ gas

$=\frac{10}{32}\times 6.023\times {10}^{23}\text{ molecules of }{\mathrm{O}}_2$

$=1.88\times {10}^{23}\text{ molecules of }{\mathrm{O}}_2$