# Question 67e63

Aug 30, 2015

The block has a mass of $\text{33,000 g}$.

#### Explanation:

In order to be able to solve this problem, you need to know osmium's density, which is listed at ${\text{22.59 g/cm}}^{3}$.

Now, your block can be thought of as a rectangular prism, which has a volume of

$\textcolor{b l u e}{{V}_{\text{rectangular prism" = "length" xx "height" xx "width}}}$

Since osmium's density is given in grams per cubic centimeters, you'll need to convert your dimensions from inches to centimeters by using the fact that $\text{1 inch" = "2.54 cm}$.

This means that you can write

5.0color(red)(cancel(color(black)("in"))) * "2.54 cm"/(1color(red)(cancel(color(black)("in")))) = 5.0 * "2.54 cm"#

You can write the volume of the block as

$V = 5.0 \cdot \text{2.54 cm" * 4.0 * "2.54 cm" * 4.5 * "2.54 cm}$

$V = 5.0 \cdot 4.0 \cdot 4.5 \cdot 2.54 {\text{^3 "cm}}^{3}$

$V = {\text{1475 cm}}^{3}$

Density is defined as mass per unit of volume

$\textcolor{b l u e}{\rho = \frac{m}{V}}$

This means that you can determine the mass of the block by

$m = \rho \cdot V = 22.57 \text{g"/color(red)(cancel(color(black)("cm"^3))) * 1475color(red)(cancel(color(black)("cm"^3))) = "33291 g}$

You need to round this off to two sig figs, the number of sig figs you have for the dimensions of the block

$m = \textcolor{g r e e n}{\text{33,000 g}}$