Question

A sample of hydrogen gas was collected over water at $21^{\circ} C$ and $685\, mm\, Hg$. The volume of the container was $7.80\, L$. Calculate the mass of $H_{2}(g)$ collected (vapour pressure of water $=18.6 \,mm Hg$ at $21^{\circ} C$ )

• A. 0.283 g
• B. 0.570 g
• C. 0.589 g
• D. 7.14 g

Answer 1

Answer: (B)

The volume of $H_{2}$ at STP
$V_{2}=\frac{P_{1} V_{1}}{T_{1}} \times \frac{T_{2}}{P_{2}}$
Here vapour pressure of $H_{2}$ at $21^{\circ} C$
$=685\, mm -18.6 \,mm =666.4 \,mm$
So $V _{2}=\frac{666.4\, mm \,Hg \times 7.80 \,L \times 273 \,K }{294 \,K \times 760 \,mm }=6.38\, L$
At STP $22.4\, L \,H _{2}$ weight $=2 g$
So, at STP $6.38 \,L\, H _{2}$ weigh
$=\frac{2\, g \times 6.38 \,L }{22.4\, L }=0.570 \,g$