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A sample of Hydrogen Gas was collected own water at 2 1^{\circ} \mathrm{C} and 600 \mathrm{MM} \mathrm{Hg}.
The volume of the container was 5 \mathrm{~L}. calculate the mass of \mathrm{H}_2(\mathrm{~g}) collected

[vapour buessume of water = 18.6 \mathrm{MM} \mathrm{Hg} \mathrm{at} 21^{\circ} \mathrm{C} \text { ] }
 

Option: 1

0.570 g


Option: 2

0.3125 g


Option: 3

0.71 g


Option: 4

0.283 g


Answers (1)

\text { The volume of } \mathrm{H}_2 \text { at STP }

V_2=\frac{P_1 V_1}{T_1} \times \frac{T_2}{P_2}

\text { Here vapour pressure of } \begin{aligned} \mathrm{H}_2 \text { at } 21^{\circ} \mathrm{C} & =600-18.6 \\ & =581.4 \mathrm{MM} \end{aligned}

so,

\begin{aligned} v_2 & =\frac{581.4 \mathrm{MMHg} \times 5 \mathrm{~L} \times 273}{294 \mathrm{~K} \times 760 \mathrm{MMHg}}=\frac{793611}{223440} \\ & =3.5 \mathrm{~L} \end{aligned}

\text { At STP } 22.4 \mathrm{~L} \mathrm{H}_2 \text { weight }=2 \mathrm{~g}

\text { so at STP }

3.5 \mathrm{~L} \mathrm{H2} \text { wight }=\frac{2 \times 3.5}{22.4}=.3195 \mathrm{~g}

 

Posted by

Ramraj Saini

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