# What is the molarity of a solution formed from 7.1 moles BaCl_2 dissolved into .85 L of water?

Over $8$ $m o l \cdot {L}^{-} 1$.
Anyway, $\text{Concentration}$ $=$ $\text{Moles of solute (moles)"/"Volume of solution (Litres)}$ $=$ $\frac{7.1 \cdot m o l}{0.85 \cdot L}$ $=$ ??*mol*L^-1
The given concentration is with respect to $\text{moles of barium sulfate per litre}$.