# What is rho_"titanium" in lb*"inch"^-3?

Sep 8, 2016

You want the density of titanium of metal in $l b \cdot {\text{inch}}^{-} 3$?

#### Explanation:

${\rho}_{\text{titanium metal}}$ $=$ $4.51 \cdot g \cdot c {m}^{-} 3$

There are $454 \cdot g \cdot l {b}^{-} 1$, and $2.54 \cdot c m \cdot {\text{inch}}^{-} 1$.

${\rho}_{\text{titanium metal}}$ $=$ $\text{Mass"/"Volume}$

$=$ $4.51 \cdot \cancel{g} \times \frac{1}{454 \cdot \cancel{g} \cdot l {b}^{-} 1} \times \frac{1}{\cancel{c {m}^{3}} \times {\left(2.54 \cdot \cancel{c m} \cdot {\text{inch}}^{-} 1\right)}^{3}}$

$=$ $0.163 \cdot l b \cdot {\text{inch}}^{-} 3$??

I think I've got this right; obviously you won't build a titanium structure on my say so! I would be more sanguine about my answer in $g \cdot c {m}^{-} 3$.