Chemistry - Mole Concept Question with Solution | TestHub

At$25°\mathrm{C}$, the water so-obtained is in the liquid state, which is supposed to occupy negligible volume or almost zero volume.

Total volume of${\mathrm{H}}_2$and${\mathrm{O}}_2$before reaction
$=100\mathrm{m}\mathrm{L}+500\mathrm{m}\mathrm{L}=600\mathrm{m}\mathrm{L}$
and volume of${\mathrm{H}}_2$and${\mathrm{O}}_2$reacted = 200 mL + 100 mL = 300 mL
Volume of${\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$formed = 0.00 mL
So decrease in volume$=(600-300+0.00)\mathrm{m}\mathrm{L}=300\mathrm{m}\mathrm{L}$