Question #ef6e4

1 Answer
Apr 23, 2017

Answer:

Add 100 mL of the concentrated #"HCl"# to enough water to make 2 L of solution.

Explanation:

In your dilute solution, you want to have

#2 color(red)(cancel(color(black)("L HCl"))) × "0.6 mol HCl"/(1 color(red)(cancel(color(black)("L HCl")))) = "1.2 mol HCl"#

Assume that you have 1 L of the concentrated #"HCl"# solution.

#"Mass of solution" = 1000 color(red)(cancel(color(black)("mL solution"))) × "1.2 g solution"/(1 color(red)(cancel(color(black)("mL solution")))) = "1200 g solution"#

#"Mass of HCl" = 1200 color(red)(cancel(color(black)("g solution"))) × "30 g HCl"/(100 color(red)(cancel(color(black)("g solution")))) = "360 g HCl"#

#"Moles of HCl" = 360 color(red)(cancel(color(black)("g HCl"))) × "1 mol HCl"/(36.46 color(red)(cancel(color(black)("g HCl")))) = "9.9 mol HCl"#

So, 1 L of the concentrated solution contains 9.9 mol #"HCl"#.

You want to get 1.2 mol #"HCl"#.

#"Volume of concentrated HCl" = 1.2 color(red)(cancel(color(black)("mol HCl"))) × "1000 mL HCl"/(9.9 color(red)(cancel(color(black)("mol HCl")))) = "100 mL HCl"#

Note: The answer can have only one significant figure, because that is all you gave for the volume and the molarity of the dilute #"HCl"#.