# Question #19c82

We need a $14.9 \cdot m L$ volume of the stock solution.
We require $0.425 \cdot L \times 0.210 \cdot m o l \cdot {L}^{-} 1 = 0.08925 \cdot m o l$.
If there is $6.0 \cdot m o l \cdot {L}^{-} 1$ stock solution available, we need a volume of $\frac{0.08925 \cdot m o l}{6.0 \cdot m o l \cdot {L}^{-} 1} \times {10}^{3} \cdot m L \cdot {L}^{-} 1 = 14.9 \cdot m L$. This volume is then diluted to $425 \cdot m L$.