# What volume of 1.0 M CuSO_4 would you need to make .750 L of a .20 M solution?

May 13, 2016

$0.150 \cdot L$ are required.

#### Explanation:

$\text{Concentration}$ $=$ $\text{Moles of solute"/"Volume of solution}$.

And thus $\text{Moles of solute } \left(n\right)$ $=$ $\text{Concentration "(mol*L^-1)xx"volume } \left(L\right)$

In the required volume there are $0.750 \cdot L \times 0.20 \cdot m o l \cdot {L}^{-} 1$ $=$ $0.150 \cdot m o l$.

Thus, in the starting volume, we need $0.150 \cdot L$, which contains $0.150 \cdot m o l$. This volume is then diluted to $0.750 \cdot L$.