Question

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`?

Answer 1

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?

Explanation:

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`