00:01
Before we do any calculations in this problem, let's note the number of ions that we have in this structure.
00:07
For our chloride ions, we have 1 eighth times 8, and we'll have a total of 1 chloride ion.
00:14
And for cesium, we have, again, one ion.
00:18
So we can find the mass of our cell by dividing the mass of cesium by the mass of chlorine.
00:26
And when we do this, we're actually going to add these values, because we are going to divide.
00:31
By avogadro's number.
00:33
So we'll take 132 .9 grams per mole, divide that by avogadro's number, and we're going to add 35 .45 grams per mole and divide that number by avogadro's number.
00:48
And when we do this addition, we'll get that our mass of our cell is 27 .95 times 10 to the 23rd grams.
00:58
So we were told that the density that we are dealing with is 3 .97 grams per centimeter cubed.
01:09
So by rearranging the density formula, we can find that volume is equivalent to mass divided by density.
01:16
So we can take the mass we just calculated and divide that by our density, and we'll get that the volume of our cell, the 7 .04 times 10 to the negative 23rd centimeters cubed.
01:35
So remember that the edge length of the unit cell is denoted by a...