# What is the molarity of a solution prepared by combining 125 mL of 0.251 molar HCl with 250 mL of pure water?

##### 1 Answer

#### Explanation:

So, you know that you're mixing *pure water*.

In essence, you're diluting the initial solution by keeping the number of moles of solute, which in your case is hydrochloric acid, **constant** and increasing the amount of *solvent*.

This means that you can expect the molarity of the second solution to be **smaller** than that of the initial solution, which would be characteristic of a *diluted* solution.

Use the volume and the molarity of the initial solution to determine how many moles of hydrochloric acid you get in that volume

#color(blue)(c = n/V implies n = c * V)#

#n_"HCl" = "0.251 M" * 125 * 10^(-3)"L" = "0.031375 moles HCl"#

The **total volume** of the final solution will be equal to

#V_"total" = "125 mL" + "250 mL" = "375 mL"#

The molarity of the second solution will thus be

#c = "0.031375 moles"/(375 * 10^(-3)"L") = color(green)("0.0874 M")#

The answer is rounded to two sig figs, the number of sig figs you have for the volume of the target solution.