# What is the final concentration of a solution prepared by adding water to 50.0 mL of 1.5 M NaOH to make 1.00 L of solution?

Moles/volume of solution = $\frac{50.0 \times {10}^{- 3} \cancel{L} \times 1.50 \cdot m o l \cdot \cancel{{L}^{- 1}}}{1.00 \cdot L}$
Given concentration is dimensionally consistent. The answer has units of $m o l \cdot {L}^{-} 1$ as required.