00:02
So continuing on with unit cells, we're specifically looking at nacl unit cell where we have na plus add cl minus in equal parts because they both have a plus or minus one charge.
00:17
So we have four na plus and four cl minus.
00:27
According to the information we have, there is an absence of one of the sodium cations and one chlorine anion.
00:37
So one nacl unit is missing from our structure.
00:41
Therefore, in nacl structure, there are three sodium cations and three chloride anions.
00:52
Therefore, the overall stoichiometry will not change and it would be the exact same.
00:56
So therefore, the density of the ions can be written as follows.
01:18
Similarly, the density of nacl structure is the following.
01:23
D2 divided by volume.
01:32
So the change in density is calculated as far as.
01:35
Follows.
01:38
We have p1, d1, take away p2, where these p symbols are representing our density, not any sort of pressure, equals three times mass over n, a over v, subtract four times m over n, a over v, all over four times m over n a over v.
02:04
This is equal to negative 1 over 4, therefore, is a reduction in 25%.
02:25
So in the second part now, we have an ag plus cat -ions, a silver cat -ion is changed.
02:37
It's movement in our ag -cl structure, so that's silver chloride, but we're not missing the ions, so the structure remains the same as n -a -c -l...