# 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?

Sep 22, 2016

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

#### Explanation:

If there are $2.55 \times {10}^{24}$ $\text{titanium atoms}$, there are $\frac{2.55 \times {10}^{24}}{6.022 \times {10}^{23} \cdot m o {l}^{-} 1}$ $=$ $4.23 \cdot m o l$ $\text{titanium atoms}$.

And this has a mass of $4.23 \cdot m o l \times 47.87 \cdot g \cdot m o {l}^{-} 1$ $=$ $202.5 \cdot g$.

Now we have a cube that has a mass of $202.5 \cdot g$. But we know its $\text{density}$, which you have kindly provided, $\rho$ $=$ $4.50 \cdot g \cdot c {m}^{-} 3$.

$\rho$ $=$ $\text{Mass"/"Volume}$. Thus $\text{Volume}$ $=$ $\frac{\text{Mass}}{\rho}$ $=$ $\frac{202.5 \cdot g}{4.50 \cdot g \cdot c {m}^{-} 3}$ $=$ $45.0 \cdot c {m}^{3}$

And thus the side-length of the metal cube $=$ ""^3sqrt(45.0*cm^3) $=$ ??cm $\cong$ $3.5 \cdot c m$

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