# What volume of 0.565 M K_2CO_3 contains exactly 0.115 moles K_2CO_3?

Apr 12, 2017

$V = 0.204$ $L$ or $20.4$ $m L$

#### Explanation:

The units of molarity are $\frac{m o l}{l i t r e}$, using this information, we can solve for the volume.

$\frac{L}{0.565 m o l} \cdot 0.115 m o l$

$\frac{L}{0.565 \cancel{m o l}} \cdot 0.115 \cancel{m o l}$

$V = \frac{0.115}{0.565}$

$V = 0.204$ $L$ or $20.4$ $m L$