# How would you relate the volume of a container (in cm3) to the volume of liquid it contains (in mL)?

Sep 3, 2016

Both the $c {m}^{3}$ and the $m L$ are units of volume. It is a fact that there are ${10}^{3} \cdot L$ in a ${m}^{3}$. How does this help us?

#### Explanation:

$1 \cdot m L = 1 \times {10}^{-} 3 L$

But $1 \cdot c {m}^{3}$ $=$ ${\left(1 \times {10}^{-} 2 m\right)}^{3}$ $=$ $1 \times {10}^{-} 6 \cdot {m}^{3}$

$=$ ${10}^{-} 6 \cdot {m}^{3} \cdot {L}^{-} 1 \times {10}^{3} \cdot L {m}^{-} 3$ $=$ ${10}^{-} 3 \cdot L$ $=$ $1 \cdot m L$ as required.

Sometimes, especially in English (i.e. UK) texts, you will find the $L$ quoted as $1 \cdot {\mathrm{dm}}^{3}$, i.e $\text{deci =} \frac{1}{10}$.

$1 \cdot {\mathrm{dm}}^{3}$ $=$ ${\left(1 \times {10}^{-} 1 m\right)}^{3}$ $=$ $\frac{1}{10} ^ 3 {m}^{3}$ $=$ $1 \cdot L$ as required.