density of Magnesium!
posted by sandy .
Magnesium crstallizes in the hcp arrangment. The dimensions of the unit cell are height 520om, length on an edge, 320. Calculate the density of Mg (s) and compare it with the measured value of 1.738 g/cm^3
this is what i did :
m= 2* 24.30g/6.022*10^23 = 8.07*10^23
v= l^2*h
v= (320)^2*520= .0053248
d=m/v
d= 8.07*10^23/. 0053248
d= 1.51* 10^20 cm
BUT the answer is suppose to be 1.75
I think I made a mistake on the volume :S

You don't list units for all of the measurements but I assume the 320 and 520 are picometers. You must convert the volume you have, which I assume is in cubic picometers, OR convert to cm before you begin. I highly recommend that you convert 320 to cm and convert 520 to cm.
320 pm x (1 m/10^12 pm) x (100 cm/1 m) = 3.2 x 10^8 cm BEFORE doing the calculations. See if that helps. 
sorry yes they were in pm I corrected my units:
m= 2* 24.30g/6.022*10^23 = 8.07*10^23
v= l^2*h
v=(3.2*10^8cm)^2* 5.2*10^8cm v=5.32*10^23cm
d=m/v
d= 8.07*10^23/5.32*10^23
d= 1.52cm
> it's still 1.52 where as the correct answer is suppose to be 1.75g/cm^3. Is my method of calculation correct? 
Is your volume formula consistent with the hexagonal close packed crystal structure? That might be the cause of your disagreement.
See http://www.chm.davidson.edu/ChemistryApplets/Crystals/UnitCells/hcp.html 
that just pretty much explains how to calculate height. and the length = 2r

It says a lot more than that. The base area of a unit cell is not the square of 2r. It is a hexagon

To find the volume of a HCP structure given the height and the lenght, you should use this formula:
l^2*h*((3)^1/3)/2)
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