Question

How do you calculate the molar volume of a gas at STP?

Answer 1

Hint: Molar volume is the volume occupied by one mole of a gas and it is measured at standard temperature and pressure. It can be calculated by dividing the molar mass of gas for which molar volume is to find out, dividing it by its mass density. The answer obtained is the molar volume and the units are $mol\,{L^{ - 1}}$ .

Complete solution:
Here we have to understand firstly that what is volume of a gas, simply it can be defined as the amount of particles present in \[1\,{m^3}\] , now we can understand that molar volume is the volume of $1\,mole$ chemical compound, here as we are talking about a gas so molar volume is the volume occupied by $1\,mole$ of gas.
We get to know about the molar volume of a gas, now a question is arising in your mind that what STP is. STP is standard temperature and pressure conditions, according to these the temperature should be $273\,K\,or\,{0^ \circ }C$ and pressure should be $1\,atm\,\,or\,766\,mm\,Hg$ . The volume measure at these conditions is called molar volume at STP, which can be represented as ${V_m}$ .
We can calculate molar volume by dividing the molar mass by mass density. An important thing to keep in mind is that at a specific temperature and pressure volume is fixed, it only changes when we change the temperature and pressure conditions for it.
${V_m} = \dfrac{{Molar\,mass}}{{Density}}$
Experimentally, one mole of any gas occupies a volume of 22.4 litres at STP. The equation can be expressed as
$1\,mole\,of\,gas\,at\,STP\, = \,22.4\,litres\,of\,gas$
Let’s see an example for your better understanding of this concept, if we want to calculate molar volume for a sample of a nitrogen gas, if its density is given as $1.250\,g\,{L^{ - 1}}$ , we have to find molar volume of nitrogen gas. Here, on applying the above formula for molar volume of nitrogen, it is then written as per-
${V_m}({N_2}) = \dfrac{{Molar\,mass\,of\,nitrogen\,gas}}{{Density\,of\,nitrogen\,gas}}$
Now if we put the molar mass of nitrogen gas, for one atom mass is $14\,g\,mo{l^{ - 1}}$ and for ${N_2}$ it will be $28\,g\,mo{l^{ - 1}}$, the density of nitrogen gas is given as $1.250\,g\,{L^{ - 1}}$ .
Now putting the both values in the equation -
${V_m}({N_2}) = \dfrac{{28\,g\,mo{l^{ - 1}}\,}}{{1.250\,g\,{L^{ - 1}}}}$ = $22.408\,L\,mo{l^{ - 1}}$

Note:Write unit after dividing the molar volume by its mass density. The answer we get is written with the unit. The unit of molar volume is $L\,mo{l^{ - 1}}$ as it is the volume of one mole of a gas and written by dividing the molar mass with the density term. Some units get cancelled and we get $L\,mo{l^{ - 1}}$ at the end. $Unit\,of\,{V_m}({N_2}) = \dfrac{{\,g\,mo{l^{ - 1}}\,}}{{g\,{L^{ - 1}}}} = \,\dfrac{{\,mo{l^{ - 1}}\,}}{{\,{L^{ - 1}}}} = L\,mo{l^{ - 1}}$