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

1 Answer
Apr 12, 2017

V=0.204 L or 20.4 mL

Explanation:

The units of molarity are (mol)/(litre), using this information, we can solve for the volume.

L/(0.565 mol)*0.115 mol

L/(0.565 cancel(mol))*0.115 cancel(mol)

V=0.115/0.565

V=0.204 L or 20.4 mL