Chemistry - Ideal Gas Question with Solution | TestHub

The volume of${\mathrm{H}}_2$at STP
${\mathrm{V}}_2=\frac{{\mathrm{P}}_1{\mathrm{V}}_1}{{\mathrm{T}}_1}\times \frac{{\mathrm{T}}_2}{{\mathrm{P}}_2}$
Here vapour pressure of${\mathrm{H}}_2$at$21°\mathrm{C}$
= 685 mm – 18.6 mm = 666.4 mm
So${\text{V}}_2=\frac{666.4\text{mm}\text{Hg}\times \text{7}\text{.80}\text{L}\times \text{273}\text{K}}{294\text{K}\times \text{760}\text{mm}}=6.38\text{L}$
At STP$\text{22}\text{.4}\text{L}{\text{H}}_{\text{2}}\text{weight}\text{=}\text{2}\text{g}$
So, at STP$\text{6}\text{.38}\text{L}{\text{H}}_{\text{2}}\text{weigh}\text{=}\frac{\text{2}\text{g}\text{×}\text{6}\text{.38}\text{L}}{\text{22}\text{.4}\text{L}}\text{=}\text{0}\text{.570}\text{g}$