If the radius Na^+ is 1A and Cl^- is 1.8A then we can assume the bond is 2.8A long and that is the length of one side of the cube of NaCl. We assume there is no space between the ions.
So the volume of the cube will be (2.8E-8 cm)^3 = 2.195E-24 cc.
How much does one molecule of NaCl weigh? That will be 58.44/6.02E23 = ? g mass.
So density = mass/volume = 2.195E-23/9.707E-23 = about 4.4 g/cc and compare that to the measured value of 2.17 g/cc.
The ionic radii of Na+ (atomic weight 22.99) and Cl- (atomic weight 35.45) are 1.0 and 1.8 Angstrom, respectively. Treating the ions as hard spheres, calculate the density of NaCl. Compare this with the measured density of 2.17 g/cm3.
1 answer
1 answer