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A sample of hydrogen gas was collected over water at 20^{\circ} \mathrm{C} and 600 \, \mathrm{MM\, Hg}.
The volume of container 9 L. 
caloulate mass at \mathrm{H}_{2}(\mathrm{g} ) collected
[vapour presswe af water \mathrm{=18.6 \mathrm{Mm} \, \mathrm{Hg}\, \mathrm{at}\, 20^{\circ} \mathrm{C}}]

Option: 1

570 g


Option: 2

283 g


Option: 3

7.14 g


Option: 4

6.58 g


Answers (1)

best_answer

volume of \mathrm{H}_{2} at S T P,

\mathrm{v_{2}=\frac{P_{1} v_{1}}{T_{1}} \times \frac{T_{2}}{P_{2}}}
Here vapour pressure at \mathrm{H_{2}} at \mathrm{20^{\circ} \mathrm{C}=600-18.6}
                                                              \mathrm{=581.4\, \mathrm{MM}}

So
\mathrm{v_{2}=\frac{581.4 \times 9 \times 273}{294 \times 760}=6.3}

\mathrm{At \, STP 22.4 \mathrm{~L} \mathrm{H}_{2} weight =2 \mathrm{g} }
so  at  STP

\mathrm{6.3 \mathrm{~L} \mathrm{H}_{2} \text { weight }=\frac{2 \times 6.3}{22.4}=570 \mathrm{~g}}

Posted by

Gautam harsolia

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