Question

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

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

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 $6.38 \, L \, H_{2} \, weigh= \, \frac{2 g \times 6 . 38 \, L}{22 . 4 \, L}=0.570 \, g$