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  • A sample at hydrogen gas was collected with water at and <img alt="\mathrm{600\, mm\, H_{g}}" src

A sample at hydrogen gas was collected with water at 21^{\circ}C and \mathrm{600\, mm\, H_{g}}.
The volume at the container was 5L.
Calculate the mass at \mathrm{H_{2}\left ( g \right )} collected
[vapour Pressure at water \mathrm{=18.6\, \mathrm{MM}\, \mathrm{Hg}\: \mathrm{at} \: 21^{\circ} \mathrm{C}} ]

Option: 1

-570 g


Option: 2

-3125 g


Option: 3

71 g


Option: 4

283 g


Answers (1)

The volume of \mathrm{H}_{2}  at STP
\mathrm{V_{2}=\frac{P_{1} V_{1}}{T_{1}} \times \frac{T_{2}}{P_{2}}}
\mathrm{\text { Here vapour possum of } \mathrm{H}_{2} \text { at } 21 \degree \mathrm{C} =600-18.6 \\ }
                                                                              \mathrm{=581.4 \, \mathrm{MM} }

So,
\mathrm{v_{2} =\frac{581.4 \mathrm{MMHg} \times 5 \mathrm{~L} \times 273}{294 \mathrm{~K} \times 760 \mathrm{\, MM\, Hg}}=\frac{793611}{223440} }
   \mathrm{=3.5 \mathrm{~L} }

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

So at STP

\mathrm{3.5 \mathrm{L} \mathrm{H} \text {, wight }=\frac{2 \times 33.5}{22.4}=3125 \mathrm{~g}}
                                                  

Posted by

Ramraj Saini

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