# What molar quantity of potassium bromide is present in a 68.5*mL volume of a solution whose concentration is 1.65*mol*L^-1 with respect to the salt?

Aug 11, 2016

Approx. $0.1$ $m o l$ $K B r$
$\text{Molarity}$ $=$ $\text{Moles of solute"/"Volume of solution}$.
Thus $\text{moles of KBr}$ $=$ $68.5 \times {10}^{-} 3 \cancel{L} \times 1.65 \cdot m o l \cdot \cancel{{L}^{-} 1}$ $=$ $0.113 \cdot m o l$.
I assume (perhaps wrongly!) that you quoted a $\text{molar}$ concentration (i.e. $m o l \cdot {L}^{-} 1$) with respect to $K B r$. You might have specified a $\text{molal concentration}$, i.e. $\text{moles of solute"/"kilogram of solvent}$, which is what "$\text{m}$" generally signifies. The answers would be similar.