Question

A mixture of \(0.200\;{\rm{g}}\) of \({{\rm{H}}_2},1.00\;{\rm{g}}\) of \({{\rm{N}}_2}\), and \(0.820\;{\rm{g}}\)of \({\rm{Ar}}\)is stored in a closed container at STP. Find the volume of the container, assuming that the gases exhibit ideal behaviour.

Short answer

The volume of the container is 3.48 litres.

Step-by-step solution

Definition of volume

A substance's volume is the amount of space it occupies.

Converting gas into number of moles

Converting \(0.2\) grams of \({{\bf{H}}_2}\) gas into no. of moles of \({{\bf{H}}_2}\)gas.

Atomic weight of Hydrogen \((H) = 1\)grams.

So, molecular weight of \({{\rm{H}}_2} = (2) \cdot (1)\)grams.

\( = 2\)grams.

i.e. \(2\) grams of \({{\rm{H}}_2} = 1\) mole of \({{\rm{H}}_2}\).

So, no. of moles of gas of \({{\rm{H}}_2}\left( {{n_1}} \right) = \left( {0.2} \right.\) grams of \(\left. {{{\rm{H}}_2}} \right) \times \left( {\frac{{1{\rm{ mole }}}}{{2{\rm{ grams of }}{{\rm{H}}_2}}}} \right) = 0.1\) moles of \({{\rm{H}}_2}\).

Converting 1 grams of \({{\bf{N}}_2}\) gas into no. of moles of \({{\bf{N}}_2}\) gas.

Atomic weight of Nitrogen \((N) = 14\) grams.

So, molecular weight of \({{\rm{N}}_2} = (2) \cdot (14)\) grams.

\( = 28\)grams.

i.e. \(28\) grams of \( = {{\rm{N}}_2} = 1\) mole of \({{\rm{N}}_2}\).

So, no. of moles of gas of \({{\rm{N}}_2}\left( {{n_2}} \right) = \left( 1 \right.\) grams of \(\left. {{{\rm{N}}_2}} \right) \times \left( {\frac{{1{\rm{ mole }}}}{{28{\rm{ grams of }}{{\rm{N}}_2}}}} \right)\)

\( = 0.035\)moles of \({{\rm{N}}_2}\)

Find the volume of the container

Converting \(0.82\) grams of \(Ar\) gas into no. of moles of \(Ar\)gas.

Molecular weight of Argon \(({\rm{Ar}}) = 39.94\) grams.

i.e. \(39.94\) grams of \(Ar\)\( = 1\)mole of \(Ar\).

So, no. of moles of gas of \({\mathop{\rm Ar}\nolimits} \left( {{n_3}} \right) = (1\) grams of \({\rm{Ar}}) \times \left( {\frac{{1{\rm{ mole }}}}{{39.94{\rm{ grams of Ar}}}}} \right)\)

\( = 0.0205\)moles of \(Ar\).

Total no. of moles \( = (0.1) + (0.035) + (0.0205)\)moles. \( = 0.1555\) moles.

Under STP conditions, Volume occupied by \(1\) mole of gas \( = 22.4\)Liters.

Therefore, for \(0.1555\)moles, Volume occupied at \(STP\)condition is

\(\begin{aligned}{} &= (0.155) \cdot (22.4){\rm{ Liters}}{\rm{.}}\\ &= 3.48{\rm{ Liters}}{\rm{.}}\end{aligned}\)