# How many grams of HCl are required to prepare 250 milliliters of a 0.158 M solution?

Jun 5, 2016

$\text{1.44 g}$

#### Explanation:

The problem gives you the molarity and volume of the target solution, so right from the start you know that you can use that information to determine the number of moles of hydrochloric acid that it must contain.

So, molarity measures a solution's concentration in terms of how many moles of solute, which in your case is hydrochloric acid, you get per liter of solution.

A $\text{0.158 M}$ hydrochloric acid solution contains $0.158$ moles of hydrochloric acid for very liter of solution. Since you're dealing with $\text{250 mL}$, the equivalent of $\frac{1}{4} \text{th}$ of a liter, your solution will contain

250 color(red)(cancel(color(black)("mL"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * "0.158 moles HCl"/(1color(red)(cancel(color(black)("L")))) = "0.0395 moles HCl"

Now that you know how many moles of hydrochloric acid are needed to prepare your solution, use the compound's molar mass to convert this to grams

0.0395 color(red)(cancel(color(black)("moles HCl"))) * "36.46 g"/(1color(red)(cancel(color(black)("mole HCl")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("1.44 g")color(white)(a/a)|)))

I'll leave the answer rounded to three sig figs.