# Question #9c63a

$0 , 571 M$
Since concentration is moles per unit volume $\left(c = \frac{n}{V}\right)$, and moles is mass over molar mass $\left(n = \frac{m}{{M}_{r}}\right)$, we may find the molar concentration (molarity) as follows :
$c = \frac{n}{V} = \frac{6 m o l}{10 , 5 {\mathrm{dm}}^{3}} = 0 , 571 m o l / {\mathrm{dm}}^{3}$