# Question #fdd32

Sep 5, 2017

$430. g$

#### Explanation:

This is simple unit conversion!

Start by converting the density to something favorable:

$\frac{1.11 g}{c {m}^{3}} \cdot \frac{c {m}^{3}}{m L} = \frac{1.11 g}{m L}$

Now things don't seem confusing, do they?

$387 m L \cdot \frac{1.11 g}{m L} \approx 430. g$

The period indicates three significant figures, what is required in this problem.