# 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?

Sep 11, 2017

#### Answer:

Well, we know that ${N}_{A}$ titanium atoms have a mass of $47.87 \cdot g$. Did I know that number from the top of my head? From where did I get it?

#### Explanation:

And since ${N}_{A} = 6.022 \times {10}^{23} \cdot m o {l}^{-} 1$, 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 $\rho ,$ $\text{density"="Mass"/"Volume}$

But the volume of a cube is ${r}^{3}$, where $r$ is the length of the sides of the cube.....

And so $\rho = \text{Mass"/"Volume}$, $\frac{\text{Volume"=r^3="Mass}}{\rho}$

$r = {\text{^(3)sqrt("Mass"/"Density")=}}^{3} \sqrt{\frac{24.64 \cdot g}{4.5 \cdot g \cdot c {m}^{-} 3}} = 1.76 \cdot c m$