A cube of titanium metal contains #3.10xx10^23"titanium atoms"#. What is the edge length of the cube if #rho_"titanium"=4.5g*cm^3#?

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Answer 1 · 2017-09-11T05:07:43

Well, we know that #N_A# titanium atoms have a mass of #47.87*g#. Did I know that number from the top of my head? From where did I get it?

And since #N_A=6.022xx10^23*mol^-1#, we take the quotient.....

#"Mass of Ti"=(3.10xx10^23*"Ti atoms")/(6.022xx10^23*"Ti atoms"*mol^-1)xx47.87*g*mol^-1=24.64*g..........#

And we know that #rho,# #"density"="Mass"/"Volume"#

But the volume of a cube is #r^3#, where #r# is the length of the sides of the cube.....

And so #rho="Mass"/"Volume"#, #"Volume"=r^3="Mass"/rho#

#r=""^(3)sqrt("Mass"/"Density")=""^(3)sqrt((24.64*g)/(4.5*g*cm^-3))=1.76*cm#

A cube of titanium metal contains #3.10xx10^23*"titanium atoms"#. What is the edge length of the cube if #rho_"titanium"=4.5*g*cm^3#?