# Calculate the molarity of 1.5% w/v solution of PbCl2?

We ASSUME that a $100 \cdot m L$ volume of SOLUTION contains a $1.5 \cdot g$ mass of solute..........
And thus $\text{molarity}$ $=$ $\text{Moles of solute"/"Volume of solution}$
$= \frac{\frac{1.5 \cdot g}{278.10 \cdot g \cdot m o {l}^{-} 1}}{0.100 \cdot {L}^{-} 1} = 5.4 \times {10}^{-} 2 \cdot m o l \cdot {L}^{-} 1$.