Assuming nitrogen molecule of spherical shape with the radius 2A° ,the percentage of empty space in one mole of N2gas is taken
In this question, first we will calculate the volume of one molecule of nitrogen . We are given that the nitrogen molecule is spherical in shape. Thus volume of one molecule of nitrogen will be
= ( 4/3) π r^3
where r is the radius of the sphere. We are given that the radius of one molecule of oxygen is 2 angstrom. Therefore the volume of one molecule of nitrogen will be
= ( 4/3) π ( 2 X 10^-9) dm3
Now the volume of one mole of nitrogen molecules will be = N X ( 4/3) π ( 2 X 10^-9) dm3
where N is the Avogadro's number. Thus we have
Volume occupied by one mole of nitrogen molecules = N X ( 4/3) π ( 2 X 10^-9)^3 dm3
= 6.023 X 10^23 X ( 4/3) X 3.14 ( 2 X 10^-9)^3
= 0.202 dm3
Now the volume occupied by one mole of any gas at STP is 22.4 L or 22.4 dm^3. So the ration of volume occupied by one mole of nitrogen molecules will be
= (0.202 / 22.4) X 100
= 0.901 %
So the percentage empty space will be
(100%-0.901%)=99.099%
In this question, first we will calculate the volume of one molecule of nitrogen . We are given that the nitrogen molecule is spherical in shape. Thus volume of one molecule of nitrogen will be
= ( 4/3) π r^3
where r is the radius of the sphere. We are given that the radius of one molecule of oxygen is 2 angstrom. Therefore the volume of one molecule of nitrogen will be
= ( 4/3) π ( 2 X 10^-9) dm3
Now the volume of one mole of nitrogen molecules will be = N X ( 4/3) π ( 2 X 10^-9) dm3
where N is the Avogadro's number. Thus we have
Volume occupied by one mole of nitrogen molecules = N X ( 4/3) π ( 2 X 10^-9)^3 dm3
= 6.023 X 10^23 X ( 4/3) X 3.14 ( 2 X 10^-9)^3
= 0.202 dm3
Now the volume occupied by one mole of any gas at STP is 22.4 L or 22.4 dm^3. So the ration of volume occupied by one mole of nitrogen molecules will be
= (0.202 / 22.4) X 100
= 0.901 %
So the percentage empty space will be
(100%-0.901%)=99.099%
