# How much 0.05 M HCl solution can be made by diluting 250 mL of 10 M HCl?

Feb 6, 2016

Calculate the number of MOLES of $H C l$ in the starting material. The second volume of solution depends on this molar quantity.

#### Explanation:

So, there 250xx10^-3cancelLxx 10*mol*Lcancel(""^-1) $=$ $250 \times {10}^{-} 2$ $m o l$ $H C l$.

Concentration $=$ $\frac{n}{V}$ $=$ $\text{Moles of solute"/"Volume of solution}$

We have the required concentration, $0.05$ $m o l \cdot {L}^{-} 1$. So....

$V = \frac{n}{C}$ $=$ $\frac{250 \times {10}^{-} 2 \cdot \cancel{m o l}}{0.05 \cdot \cancel{m o l} \cdot {L}^{-} 1}$ $=$ ??L