We’re being asked to **calculate the volume** of concentrated **HCl needed to prepare a 2.90 L of 0.505 M HCl.**

Recall that ** molarity** is the ratio of the moles of solute and the volume of solution (in liters).

In other words:

$\overline{){\mathbf{Molarity}}{\mathbf{}}{\mathbf{\left(}}{\mathbf{M}}{\mathbf{\right)}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}\frac{\mathbf{moles}\mathbf{}\mathbf{of}\mathbf{}\mathbf{solute}}{\mathbf{Liters}\mathbf{}\mathbf{of}\mathbf{}\mathbf{solution}}}$

We’re given the mass percent of concentrated HCl which is 37.0%.

Recall that ** mass percent** is given by:

$\overline{){\mathbf{\%}}{\mathbf{}}{\mathbf{Mass}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}\frac{\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{HCl}}{\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{solution}}}$

This means in **100 g of solution**, we have **37 g HCl**.

**The steps we need to do for this solution are:**

*Step 1:* Calculate the moles of HCl

*Step 2:* Calculate the volume of the solution (in L).

*Step 3:* Calculate the molarity of the solution.

Step 4: Calculate the volume of solution needed (in L) using dilution equation.

Hydrochloric acid is usually purchased in a concentrated form that is 37.0% HCl by mass and has a density of 1.20 g/mL.

How much concentrated solution would you take to prepare 2.90 L of 0.505 M HCl by mixing with water?

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