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

Question

Wyatt Adkins · Accepted Answer

#"1.44 g"#

You know from the outset that you can use the molarity and volume of the target solution provided by the problem to calculate the required concentration of hydrochloric acid in moles.

The number of moles of solute—hydrochloric acid in this case—that you get per liter of solution is how molarity quantifies the concentration of a solution.

A #"0.158 M"# hydrochloric acid solution contains #0.158# moles of hydrochloric acid for very liter of solution. Since you're dealing with #"250 mL"#, the equivalent of #1/4"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"#

Using the compound's molar mass, convert the number of moles of hydrochloric acid required to prepare your solution 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 round the result to three significant figures.

Xavier Avery · Answer

undefined To prepare 250 milliliters of a 0.158 M solution of HCl, you would need 9.9 grams of HCl.