#2.55xx10^24# titanium atoms are cast into a cube that has a mass of #202.5g#. If the density of titanium metal is #rho=4.60g*cm^-3#, what is the edge length of the cube?

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Answer 1 · 2016-09-22T19:20:42

The length of the metal cube #=# #""^3sqrt(45.0*cm^3)# #=# #??cm#

If there are #2.55xx10^24# #"titanium atoms"#, there are #(2.55xx10^24)/(6.022xx10^23*mol^-1)# #=# #4.23*mol# #"titanium atoms"#.

And this has a mass of #4.23*molxx47.87*g*mol^-1# #=# #202.5*g#.

Now we have a cube that has a mass of #202.5*g#. But we know its #"density"#, which you have kindly provided, #rho# #=# #4.50*g*cm^-3#.

#rho# #=# #"Mass"/"Volume"#. Thus #"Volume"# #=# #"Mass"/rho# #=# #(202.5*g)/(4.50*g*cm^-3)# #=# #45.0*cm^3#

And thus the side-length of the metal cube #=# #""^3sqrt(45.0*cm^3)# #=# #??cm# #~=# #3.5*cm#

Don't worry about making a mistake in the question; this happens. Mind you, I hope I have got your question right!

#2.55xx10^24# titanium atoms are cast into a cube that has a mass of #202.5*g#. If the density of titanium metal is #rho=4.60*g*cm^-3#, what is the edge length of the cube?