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